Atoms MCQs for NEET

Atoms

Question 1. When an α-particle of mass m moving with velocity v bombards on a heavy nucleus of charge Ze, its distance of closest approach from the nucleus depends on m as:

1. \(\frac{1}{\sqrt{m}}\)
2. \(\frac{1}{m^2}\)
3. m
4. \(\frac{1}{m}\)

Answer: 4. \(\frac{1}{m}\)

At the closest distance of approach, the kinetic energy of the particle will convert completely into electrostatic potential energy

Here, K.E=\(\frac{1}{2} m v^2\)

Electrostatic potential energy,

= \(\frac{K Q q}{d} \)

⇒ \(\frac{1}{2} m v^2 =\frac{K Q q}{d}\)

d \(\propto \frac{1}{m}\)

Question 2. In a Rutherford scattering experiment when a projectile charge Z₁ and mass M₁ approaches a target nucleus of charge Z₂ and mass M₂, the distance of the closest approach is r₀. The energy of the projectile is:

1. directly proportional to M₁ x M₂
2. directly proportional to Z₁Z₂
3. inversely proportional to Z₁
4. directly proportional to mass M₁

Answer: 2. directly proportional to Z₁Z₂

The energy of the projectile is directly proportional to Z₁ Z₂

Question 3. An alpha nucleus of energy \(\frac{1}{2} m v^2 \)bombards a heavy nuclear target of charge Ze. Then the distance of the closest approach for the alpha nucleus will be proportional to:

1. \(\frac{1}{\mathrm{Ze}}\)
2. \(v^2\)
3. \(\frac{1}{m}\)
4. \(\frac{1}{v^2}\)

Answer: 3. \(\frac{1}{m}\)

From question,\(\frac{1}{2} m v^2 =\frac{1}{4 \pi \varepsilon_0} \frac{(2 e)(z e)}{r_0}\)

⇒ \(r_0 =\frac{1}{4 \pi \varepsilon_0} \frac{2 z e^2}{\frac{1}{2} m v^2}\)

∴ \(r_0 \propto \frac{1}{m}\)

Question 4. Let T₁ and T₂ be the energy of an electron in the first and second excited states of the hydrogen atom, respectively. According to the Bohr’s model of an atom, the ratio T₁: T₂ is :

1. 1: 4
2. 4: 9
3. 4: 9
4. 9: 4

Answer: 4. 9: 4

For the first excited state, n=2

For the second excited state, n=3

∴ \(\frac{T_1}{T_2}=\frac{r_2^2}{r_1^2}=\frac{3^2}{2^2}=\frac{9}{4}\)

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Question 5. The total energy of an electron in the nth stationary orbit of the hydrogen atom can be obtained by :

1. \(\mathrm{E}_n=\frac{13.6}{n^2} e \mathrm{~V}\)
2. \(\mathrm{E}_n=-\frac{13.6}{n^2} \mathrm{eV}\)
3. \(\mathrm{E}_n=-\frac{1.36}{n^2} e \mathrm{~V}\)
4. \(E_n=-13.6 \times n^2 \mathrm{eV}\)

Answer: 2. \(\mathrm{E}_n=-\frac{13.6}{n^2} \mathrm{eV}\)

The energy of an electron in Bohr’s orbit of a hydrogen atom is given by the expression.

⇒ \(E_n=-\frac{2 \pi^2 m e^4 z^2}{n^2 h^2\left(4 \pi \varepsilon_0\right)^2}=-13.6 \frac{z^2}{n^2} \mathrm{eV}\)

Since, Z=1 for the hydrogen atom.

So, the above equation further simplifies to,

∴ \(E_n=-\frac{13.6}{n^2} \mathrm{eV}\)

Question 6. For which one of the following, the Bohr model is not valid?

1. Singly ionised helium atom (He⁺)
2. Deuteron atom
3. Singly ionised neon atom (Ne⁺)
4. Hydrogen atom

Answer: 3. Hydrogen atom

Bohr’s model is not valid for singly ionised neon K \(\left(\mathrm{Ne}^{+}\right)\).

Question 7. The total energy of an electron in an orbit is – 3.4 eV. Its kinetic and potential energies are, respectively :

1. – 3.4 eV, – 6.8 eV
2. – 3.4 eV, – 6.8 eV
3. – 3.4 eV, – 3.4 eV
4. – 3.4 eV, – 3.4 eV

Answer: 2. – 3.4 eV, – 6.8 eV

In Bohr’s model of H -atom,

⇒ \(\mathrm{K} . \mathrm{E} =|\mathrm{TE}|=\frac{|\mathrm{U}|}{2}\)

⇒ \(\mathrm{K} . \mathrm{E} =3.4 \mathrm{eV} \)

∴ \(\mathrm{U} =-6.8 \mathrm{eV}\)

Question 8. The radius of the first permitted Bohr orbit for the electron, in a hydrogen atom equals 0.51 A and its ground state energy equals – 13.6 eV. If the electron in the hydrogen atom is replaced by muon (μ^-1) [Charge same as electron and mass is 207 m_e, the first Bohr radius and ground state energy will be :

1. 0.53 x 10^-13 m, – 36 eV
2. 25.6 x 10^-13 m, – 2.8 eV
3. 2.56 x 10^-13 m, – 2.8 KeV
4. 2.56 x 10^-13 m, – 13.6 eV

Answer: 3. 2.56 x 10^-13m, – 2.8 KeV

⇒ \(r_n \propto \frac{1}{m}\)

⇒ \(r_r=\frac{0.51}{207}=2.50 \times 10^{-13} \mathrm{~m}\)

Since,\(\mathrm{E} \propto m_e\)

u=\(-13.6 \times 207=-2.8 \mathrm{KeV}\)

Question 9. The ratio of kinetic energy to the total energy of an electron in a Bohr orbit of the hydrogen atom is:

1. 2: – 1
2. 1: – 1
3. 1: 1
4. 1: – 2

Answer: 2. 1: – 1

We have that,\(\mathrm{KE}\)=- (total energy)

So kinetic energy: total energy

=1:-1

Question 10. An electron in a hydrogen atom makes a transition n₁ → n₂ where n₁ is the principal quantum number of the two states. Assuming Bohr’s model to be valid, the period of the electron in the initial state is eight times that in the final state. The possible values of n₁ and n₂ are:

1. n₁ = 6 and n₂ = 2
2. n₁ = 8 and n₂ = 1
3. n₁ = 8 and n₂ = 2
4. n₁ = 4 and n₂ = 2

Answer: 4. n₁ = 4 and n₂ = 2

The period of an electron in Bohr orbit is,

T= \(\frac{4 \varepsilon_0^2 h^3 n^3}{m e^4} \)

T \(\propto n^3 \)

⇒ \(\frac{T_1}{T_2}=\frac{n_1^3}{n_2^3} \)

As, \(T_1=8 T_2\) then

⇒ \((\frac{8 T_2}{T_2})=\frac{n_1^3}{n_2^3} \)

⇒ \((\frac{n_1}{n_2})^3=\left(2^2\right)^3\)

⇒ \(\frac{n_1}{n_2}=2^2 \)

∴ \(n_1=2 n_2 \)

Question 11. The electron in a hydrogen atom first jumps from a third excited state to the second excited state and then from the second excited to the first excited state. The ratio of the wavelengths λ₁: λ₂ emitted in the two cases is:

1. \(\frac{7}{5}\)
2. \(\frac{27}{20}\)
3. \(\frac{27}{5}\)
4. \(\frac{20}{7}\)

Answer: 4. \(\frac{20}{7}\)

We know that,

Energy, E =\(\frac{h c}{\lambda}=13.6\left[\frac{1}{h_2^2}-\frac{1}{h_1^2}\right]\)

=13.6\(\left[\frac{1}{h_1^2}-\frac{1}{h_2^2}\right]\)  → Equation 1

According to the question,

⇒ \(E_1=\frac{h c}{\lambda_1}=13.6\left[\frac{1}{3^2}-\frac{1}{4^2}\right]\)

And \(E_2 =\frac{h c}{\lambda_2} \)

=13.6\(\left[\frac{1}{2^2}-\frac{1}{3^2}\right]\) →  Equation 2

From eq. (1) and (2),

⇒ \(\frac{\lambda_1}{\lambda_2} =\frac{\frac{1}{4}-\frac{1}{9}}{\frac{1}{9}-\frac{1}{10}}=\frac{20}{7}\)

=20: 7

Question 12. An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom of mass m acquired as a result of photon emission will be: (R Rydberg constant and h Planck’s constant)

1. \(\frac{24 h R}{25 m}\)
2. \(\frac{25 h R}{24 m}\)
3. \(\frac{25 m}{24 h R}\)
4. \(\frac{24 m}{25 h R}\)

Answer: 1. \(\frac{24 h R}{25 m}\)

According to the question,

Electron passes from fifth energy level to ground level means;-

⇒ \(E_5-E_1 =\frac{h c}{\lambda} \)

⇒ \(\frac{R h c}{25}-R h c =\frac{h c}{\lambda} \)

⇒ \(\frac{24}{25} R =\frac{1}{\lambda}\)

But, p=\(\frac{h}{\lambda}\)

and v=\(\frac{h}{m \lambda}=\frac{24}{25} \frac{\mathrm{R} h}{m}\)

Question 13. Monochromatic radiation emitted when an electron on a hydrogen atom jumps from first excited to the ground state irradiates a photosensitive material. The stopping potential is measured to be 3.57 V. The threshold frequency of the material is:

1. 4 x 10¹⁵ Hz
2. 5 x 10¹⁵ Hz
3. 1.6 x 10¹⁵ Hz
4. 2.5 x 10¹⁵ Hz

Answer: 3. 1.6 x 10¹⁵ Hz

According to the question,

Energy released =(-3.4)-(-13.6)=10.2 \(\mathrm{eV}\)

Using Einstein’s photoelectric equation

W = E-e V=h v

v = \(\frac{E-e V}{h}=\frac{(10.2-3.57) e}{6.67 \times 10^{-34}} \)

v = \(\frac{6.03 \times 1.6 \times 10^{-19}}{6.67 \times 10^{-34}} \)

= 1.6 \(\times 10^{15} \mathrm{~Hz}\)

Question 14. The energy of a hydrogen atom in the ground state is – 13.6 eV. The energy of a He⁺ ion in the first excited state will be:

1. – 13.6 eV
2. – 27.2 eV
3. – 54.4 eV
4. – 6.8 eV

Answer: 1. – 13.6 eV

We know that, E=\(\frac{-13.6}{n^2} z^2\) for first excited state n=2 and for \(\mathrm{He}^{+}=2\)

E=\(\frac{-13.6}{2^2} \times 2^2=-13.6 \mathrm{eV}\)

Question 15. The electron in the hydrogen atom jumps from its excited state (n = 3) to its ground state (n = 1) and the photons thus emitted irradiate a photosensitive material. If the work function of the material is 5.1 eV, the stopping potential is estimated to be (the energy of the electron in

1. 5.1V
2. 21.1V
3. 17.2
4. 7 V

Answer: 4. 7 V

For n=1, \(E_1=\frac{13.6}{(1)^2}=-13.6 \mathrm{eV}\)

For n=3, \(E_7=\frac{13.6}{(3)^2}=-1.51 \mathrm{eV}\)

So, required energy, E=\(E_3-E_1 =-(1.51)-(-13.6)=12.09 \mathrm{eV}\)

=W+e V

e V =E-W

e V =(12.09-5.1) e

V =7 volt

Question 16. The ionization energy of the electron in the hydrogen atom in its ground state is 13.6 eV. The atoms are excited to higher energy levels to emit radiations of 6 wavelengths. Maximum wavelength:

1. n = 3 to n = 2 states
2. n = 3 to n = 1 states
3. n = 2 to n = 1 states
4. n = 4 to n = 3 states

Answer: 4. n = 4 to n = 3 states

We know that the number of spectral lines

N =\(\frac{n(n-1)}{2}\)

⇒ \(\frac{n(n-1)}{2}\) =6

⇒ \(n^2\)-n-12 =0

(n-4)(n+3) =0

n =4

Question 17. The ground state energy of the hydrogen atom is -13.6 eV. When its electron is in the first excited state, its excitation energy is:

1. 3.4 eV
2. 6.8 eV
3. 10.2 eV
4. zero

Answer: 3. 10.2 eV

Ground state energy of hydrogen atom

⇒ \(E_1=-13.6 \mathrm{eV}\)

The energy of an electron in a first excited state (n=2)

⇒ \(E_2=-\frac{13.6}{(2)^2} \mathrm{eV}\)

Excitation energy, \(\Delta E =E_2-E_1\)

= \(-\frac{13.6}{4}-(-13.6)\)

= -3.4+13.6=10.2 \(\mathrm{eV}\)

Question 18. The total energy of an electron in the ground state of a hydrogen atom is -13.6 eV. The kinetic energy of an electron in the first excited state is:

1. 6.8 eV
2. 13.6 eV
3. 1.7 eV
4. 3.4 eV.

Answer: 4. 3.4 eV.

Energy of \(n^{\text {th }}\) orbit of hydrogen atom is given by

⇒ \(E_n=-\frac{13.6}{(2)^2} \mathrm{eV}\)

for ground state n=1

⇒ \(E_1=-\frac{13.6}{12}=-13.6 \mathrm{eV}\)

for excited state n=2

∴ \(E_2=-\frac{13.6}{22}=-3.4 \mathrm{eV}\)

Question 19. The total energy of an electron in the first excited state of a hydrogen atom is about – 3.4 eV. Its kinetic energy in this state is:

1. 3.4 eV
2. 6.8 eV
3. – 3.4 eV
4. – 6.8 eV

Answer: 1. 3.4 eV

K.E.=\(\left|\frac{1}{2} P . E\right|\)

But P.E is negative,

Total energy =\(\left|\frac{1}{2} P \cdot E\right|-P . E \)

= \(-\frac{P . E}{2}=-3.4 \mathrm{eV}\)

∴ \(\mathrm{K} . \mathrm{E} =+3.4 \mathrm{eV}\)

Question 20. The Bohr model of atoms:

1. assumes that the angular momentum of electrons is quantized “
2. uses Einstein’s photo-electric equation
3. predicts continuous emission spectra for atoms
4. predicts the same emission spectra for all types of atoms

Answer: 1. assumes that the angular momentum of electrons is quantized ”

Both modes of atoms assume that the angular momentum of electrons is quantized.

∴ \(q= \pm h e\)

Question 21. When a hydrogen atom is in its first excited level, its radius is:

1. four times, it’s ground state radius
2. twice, it’s ground state radius
3. same as its ground state radius
4. half of its ground state radius

Answer: 1. four times, it’s ground state radius

The radius of \(n^{\text {th }}\) orbit of hydrogen

⇒ \(r_{\mathrm{n}}=\frac{\varepsilon_0 n^2 h^2}{\pi m e^2 Z}\)

⇒ \(r_{\mathrm{n}}=\frac{n^2 a_0}{Z} \text { or } r_{\mathrm{n}} \propto \frac{n^2}{Z}\)

For ground state, n=1

Atomic number, Z=1

For the first excited state, n=2

⇒ \(\frac{r_2}{r_1}=\left(\frac{2}{1}\right)^2=4\)

⇒ \(r_2=4 r_1\)

Hence, the radius of the first excited state is four times the radius in the ground state.

Question 22. Consider an electron in the nth orbit of a hydrogen atom in the Bohr model. The circumference of the orbit can be expressed in terms of de-Broglie wavelength \(\lambda\) of that electron as:

1. \((0.529) n \lambda\)
2. \(\sqrt{n \lambda}\)
3. \((13.6) \lambda\)
4. n \(\lambda\)

Answer: 4. n \(\lambda\)

In terms of the wavelength of the wave corresponding with an electron, the circumference of an orbit in an atom is given by,

∴ \(2 \pi r_n=n \lambda\)[\(r_n\)=radius of any n orbit]

Question 23. To explain his theory, Bohr used:

1. conservation of linear momentum
2. conservation of angular momentum
3. conservation of quantum frequency
4. conservation of energy

Answer: 2. conservation of angular momentum

Bohr used quantization of angular momentum. For stationary orbits, Angular momentum,

⇒ \(\mathrm{I} \omega(\text { omega })=\frac{n h}{2 \pi}\) .

Where,n=1,2,3,.. etc.

Question 24. If an electron in a hydrogen atom jumps from the 3^rd orbit to the 2^nd orbit, it emits a photon of wavelength \(\lambda\). When it jumps from the 4^th orbit to the 3^rd orbit, the corresponding wavelength of the photon will be:

1. \(\frac{16}{25} \lambda\)
2. \(\frac{9}{16} \lambda\)
3. \(\frac{20}{7} \lambda\)
4. \(\frac{20}{13} \lambda\)

Answer: 3. \(\frac{20}{7} \lambda\)

Using Rydberg formula we have,

⇒ \(\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \)

= \(R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=\frac{5 R}{36}\)

and \(\frac{1}{\lambda^{\prime}}=R\left(\frac{1}{3^2}-\frac{1}{4^2}\right)=\frac{7 R}{144} \)

⇒ \(\frac{\lambda^{\prime}}{\lambda}=\frac{5 R}{36}-\frac{144}{7 R}=\frac{20}{7}\)

∴ \(\lambda^{\prime}=\frac{20}{7} \lambda \)

Question 25. The transition from the state n = 3 to n = 1 in a hydrogen-like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from:

1. 2 → 1
2. 3 → 2
3. 4 → 2
4. 4 → 3

Answer: 4. 4 → 3

Infrared radiation is found in the Paschen, Bracket and p fund series and it is obtained when ‘e’ transitions from a high energy level to a minimum 3^rd energy level.

Question 26. The wavelength of the first line of the Lyman series for a hydrogen atom is equal to that of the second line of the Balmer series for a hydrogen-like ion. The atomic number Z of a hydrogen-like ion is:

1. 4
2. 1
3. 2
4. 3

Answer: 3. 2

We know that, the Lyman series for \(\mathrm{H}ion\)

⇒ \(\frac{h c}{\lambda}=\mathrm{R} h c\left(\frac{1}{1^2}-\frac{1}{2^2}\right)\)

and for H-like atoms.

⇒ \(\frac{h c}{\lambda}=\mathrm{Z}^2 \mathrm{R} h c\left(\frac{1}{2^2}-\frac{1}{4^2}\right)\)

compare we get,

⇒ \(\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=Z^2\left(\frac{1}{2}-\frac{1}{16}\right)\)

Solving we get Z=2

Question 27. An electron in the hydrogen atom jumps from the exited state n to the ground state. The wavelength so emitted illuminates a photo-sensitive material having a work function of 2.75 eV. If the stopping potential of the photo-electron is 10 V, the value of n is:

1. 3
2. 4
3. 5
4. 2

Answer: 1. 3

Using Einstein’s photoelectric equation

E =\(W+K E_{\max }=W+e V_0\)

=2.75+10=12.75 \(\mathrm{eV}\)

Difference of =4 and n = 1

the energy level is 19.75 eV

Here Higher energy level is n = 4 and the lower energy level is n = 3

Question 28. Out of the following which one is not a possible energy for a photon to be emitted by a hydrogen atom according to Bohr’s atomic model?

1. 1.9 eV
2. 11.1eV
3. 13. eV
4. 0.65 eV

Answer: 2. 11.1eV

But 11.1 eV is not present in the diagram.

So, 11.1 eV is not possible.

Question 29. The ionization potential of a hydrogen atom is 13.6 eV. Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy 12.1 eV. According to Bohr’s theory, the spectral lines emitted by hydrogen will be:

1. one
2. two
3. three
4. four

Answer: 3. three

According to question, \(12.1 =13.6\left(\frac{1}{l^2}-\frac{1}{n^2}\right)\)

⇒ \(n^2 =\frac{13.6}{1.5}\)=9

n =3

Number of spectral lines emitted

= \(\frac{n(n-1)}{2}=\frac{3 \times 2}{2}\)

= 3

Question 30. Energy levels A, B and C of a certain action correspond to increasing values of energy i.e. \(E_A<E_B<E_C\). If \(\lambda_1, \lambda_2 \) and \(\lambda_3\) are the wavelength of radiations corresponding to transitions C to B, B to A and C to A respectively, which of the following relations is correct?

1. \(\lambda_3=\lambda_1+\lambda_2\)
2. \(\lambda_3=\frac{\lambda_1 \lambda_2}{\lambda_1+\lambda_2}\)
3. \(\lambda_1+\lambda_2+\lambda_3=0\)
4. \(\lambda_3{ }^2=\lambda_1{ }^2+\lambda_2{ }^2\)

Answer: 2. \(\lambda_3=\frac{\lambda_1 \lambda_2}{\lambda_1+\lambda_2}\)

⇒ \(E_A<E_B<E_C\)

E =\(\frac{h c}{\lambda}\)

⇒ \(E_{C A} =E_{C B}+E_{B A} \)

⇒ \(\frac{h c}{\lambda_3} =\frac{h c}{\lambda_1}+\frac{h c}{\lambda_2} \)

⇒ \(\frac{1}{\lambda_3} =\frac{1}{\lambda_1}+\frac{1}{\lambda_2} \)

∴ \(\lambda_3 =\frac{\lambda_1 \lambda_2}{\lambda_1+\lambda_2}\)

Question 31. Energy E of a hydrogen atom with principal quantum number n is given by E = \(\frac{-13.6}{n^2} \mathrm{eV}\). The energy of a photon ejected when the electron jumps for n = 3 state n = 2 state of hydrogen is approximately

1. 1.5 eV
2. 0.85 eV
3. 3.4 eV
4. 1.9 eV

Answer: 4. 1.9 eV

Energy of photon=\(E_2-E_2\)

= \(\frac{13.6}{9}-\left(\frac{-13.6}{9}\right)\)

= \(\frac{5}{30} \times 13.6 \)

= 1.9 \(\mathrm{eV}\)

Question 32. When electron jumps from n = 4 to n = 2 orbit, we get:

1. the second line of the Lyman series
2. the second line of the Balmer series
3. the second line of the Paschen series
4. an absorption line of the Balmer series

Answer: 2. second line of Balmeer series

Jump to the second orbit leads to the Balmer series. When an electron jumps from the 4^th orbit to the 2^nd, orbit shall give rise to the second line of the Balmer series.

Question 33. The spectrum obtained from a sodium vapour lamp is an example of:

1. band spectrum
2. continuous spectrum
3. emission spectrum
4. absorption spectrum

Answer: 3. emission spectrum

A spectrum is noticed, when light coming straightforwardly from a source is analysed with a spectroscope. The spectrum acquired from a sodium vapour lamp is the emission spectrum.

Hence, The spectrum obtained from a sodium vapour lamp is an example of the transmission spectrum.