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Atoms NEET Questions — Free MCQ Practice

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Ratio of longest wavelengths corresponding to Lyman and Balmer series in twinge!) spectrum ts

A.

5/27

B.

3/23

C.

7/29

D.

9/31

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Explanation

(a) Wavelength for Lyman series

λ_L=1/R(1-1/4)=4/3R

and wavelength for Balmer series

λ_B=1/R(1/4-1/9)=1/R(5/36)=36/5R

∴λ_L/λ_B=4/3R x 5R/36=5/27

=>λ_L:λ_B=5:27

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Electron in hydrogen atom first jumps from third exicted state to second exicted state and then from second exicted to the first excited state. The ratio of the wavelengths $λ_{1} : λ_{2}$ emitted in the two cases is

A.

7/5

B.

27/20

C.

27/5

D.

20/7

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Explanation

Here, for wavelength $λ_{1}$

$n_{1} = 4 a n d n_{2} = 3$

and for $λ_{2}, n_{1} = 3 a n d n_{2} = 2$

We have $\frac{h c}{λ} = - 13.6 [\frac{1}{n_{2}^{2}} - \frac{1}{n_{1}^{2}}]$

So, for $λ_{1}$

$\Rightarrow \frac{h c}{λ_{1}} = - 13.6 [\frac{1}{{(4)}^{2}} - \frac{1}{{(3)}^{2}}]$

$\frac{h c}{λ_{1}} = 13.6 [\frac{7}{144}] . . . (i)$

Similarly, for $λ_{2}$

$\Rightarrow \frac{h c}{λ_{2}} = - 13.6 [\frac{1}{{(3)}^{2}} - \frac{1}{{(2)}^{2}}]$

$\frac{h c}{λ_{2}} = 13.6 [\frac{5}{36}] . . . (i i)$

Hence, from Eqs. (i) and (ii), we get

$\frac{λ_{1}}{λ_{2}} = \frac{20}{7}$

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An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquired as a result of photon emission will be

A.

$\frac{24 h R}{25 m}$

B.

$\frac{25 h R}{24 m}$

C.

$\frac{25 m}{24 h R}$

D.

$\frac{24 m}{25 h R}$

(m is the mass of the electron, R Rydberg constant and h Planck's constant)

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Explanation

$\frac{1}{λ} = R (\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}) H e r e n_{1} = 1 a n d n_{2} = 5 H e n c e \frac{1}{λ} = R (\frac{1}{1^{2}} - \frac{1}{5^{2}}) = \frac{24}{25} R N o w p h o t o n e n e r g y - E = \frac{h c}{λ} = \frac{24}{25} h c R H e r e t h e m o m e n t u m o f p h o t o n = m o m e n t u m o f a t o m T h u s P = \frac{E}{c} = \frac{24}{25} h R V e l o c i t y o f a t o m v = \frac{P}{m} = \frac{24 h R}{25 m}$

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Monochromatic radiation emitted when electron on 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:

A.

$4 \times 10^{15} H z$

B.

$5 \times 10^{15} H z$

C.

$1.6 \times 10^{15} H z$

D.

$2.5 \times 10^{15} H z$

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Explanation

Energy released from emission of electron

$E = (- 3.4) - (- 13.6)$

$= 10.2 e V$

From photo electric equation.

Work function

$ϕ = E - e V = h v$

$v = \frac{E - e V}{h}$

= $\frac{(10.2 - 3.57) e}{6.67 \times 10^{- 34}}$

v= $\frac{6.63 \times 1.6 \times 10^{- 19}}{6.67 \times 10^{- 34}}$

$= 1.6 \times 10^{15} H z$

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The transition from the state n=3 to n=1

a hydrogen like atom results in ultraviolet

radiation. Infrared radiation will be obtained

in the transition from

A.

$2 \to 1$

B.

$3 \to 2$

C.

$4 \to 2$

D.

$4 \to 3$

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Explanation

Infrared radiation is found in Paschan, Brackett

and pfund series and it is obtain when electron

transition occur from high energy level to

minimum third level.

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

A.

4

B.

1

C.

2

D.

3

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Explanation

Lyman series for H-ion

$\frac{hc}{λ} = Rhc (\frac{1}{1^{2}} - \frac{1}{2^{2}})$

and for H-like ion

$\frac{hc}{λ} = Z^{2} Rhc (\frac{1}{2^{2}} - \frac{1}{4^{2}})$

$∴ (\frac{1}{1^{2}} - \frac{1}{2^{2}}) = Z^{2} (\frac{1}{4} - \frac{1}{16}) (1 - \frac{1}{4}) = Z^{2} (\frac{1}{4} - \frac{1}{16}) Z = 2$

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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

A.

-13.6 eV

B.

-27.2 eV

C.

-54.4 eV

D.

-6.8 eV

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Explanation

Energy E of an atom with principal quantum number n is given by E= $\frac{- 13.6}{n^{2}} Z^{2}$ for first excited state n=2 and for ${He}^{+} Z = 2$

$\Rightarrow E = \frac{- 13.6 \times {(2)}^{2}}{{(2)}^{2}}$

=-13.6 eV

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

A.

$\frac{1}{Z e}$

B.

$v^{2}$

C.

$\frac{1}{m}$

D.

$\frac{1}{v^{4}}$

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Explanation

A $α ‐$ particle of mass m possesses initial velocity v, when it is at a large distance from the nucleus of an atom having atomic number Z. At the distance of closest approach, the kinetic energy of $α ‐$ particle is completely converted into potential energy. Mathematically,

$\frac{1}{2} m v^{2} = \frac{1}{4 {πε}_{0}} \frac{(2 e) (Z e)}{r_{0}}$

$r_{0} = \frac{1}{4 {πε}_{0}} \frac{2 Z e^{2}}{\frac{1}{2} m v^{2}}$

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In a Rutherford scattering experiment when a projectile of charge $Z_{1}$ and mass $M_{1}$ approaches a target nucleus of charge $Z_{2}$ and mass $M_{2},$ the distance of closest approach is $r_{0}$ . The energy of the projectile is

A.

directly proportional to $M_{1} \times M_{2}$

B.

directly proportional to $Z_{1} Z_{2}$

C.

inversely proportional to $Z_{1}$

D.

directly proportional to mass $M_{1}$

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Explanation

A particle of mass $M_{1}$ and charge $z_{1}$ possess initial velocity u, when it is at a large distance from the nucleus of an atom having atomic number $z_{2}$ . At the distance of closest approach, the kinetic energy of particle is completely converted to potential energy. Mathematically,

$\frac{1}{2} M_{1} u^{2} = \frac{1}{4 {πε}_{0}} \frac{z_{1} z_{2}}{r_{0}}$

So the energy of the particle is directly proportional to $z_{1} z_{2}$ .

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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 of emitted radiation corresponds to the transition between

A.

n=3 to n=2 states

B.

n=3 to n=1 states

C.

n=2 to n=1 states

D.

n=4 to n=3 states

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Explanation

Number of spectral lines obtained due to transition of electron from $n^{t h}$ orbit to lower orbit is

N = $\frac{n (n - 1)}{2}$ and for maximum wavelength the difference between the orbits of the series should be minimum.

Number of spectral lines N = $\frac{n (n - 1)}{2}$

$\Rightarrow \frac{n (n - 1)}{2} = 6$

or $n^{2} - n - 12 = 0$

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

or n=4

Now as the first line of the series has the maximum wavelength, therefore electron jumps from the $4^{t h}$ orbit to the third orbit.

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