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Physics NEET Questions — Free MCQ Practice

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The angular speed of electron in a hydrogen atom in $n^{th}$ orbit is proportional to -

A.

$n^{3}$

B.

$n^{- 3}$

C.

$n^{- 2}$

D.

$n^{2}$

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Explanation

$(2) ω = \frac{v}{r} ω \propto \frac{1}{n^{3}} (∵ v \propto \frac{1}{n} a n d r \propto n^{2})$

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Which of the following transition in a hydrogen atom will produce radiations of minimum wavelength?

A.

n = 2 to n = 1

B.

n = 5 to n = 3

C.

n = 10 to n = 5

D.

n = 10 to n = 3

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Explanation

The photon of higher frequency will be emitted if the transition takes place from n=2 to n=1. So the wavelength will be minimum in this transition.

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The frequency Of radiation emitted during the transition of an electron from a second excited state to a first excited state in H-atom is $f_{0} .$ The frequency of the same transition emitted by a singly ionized He ion is

A.

$2 f_{0}$

B.

$4 f_{0}$

C.

$\frac{f_{0}}{2}$

D.

$\frac{f_{0}}{4}$

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Explanation

$(2) f \propto Z^{2}$

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The energy of a hydrogen atom in its ground state is —13.6 eV. The energy of the level corresponding to the quantum number n = 2 (first excited state) in the hydrogen atom is:

A.

-2.72 eV

B.

- 0.85 eV

C.

-0.54 eV

D.

— 3.4 eV

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Explanation

$(4) E_{n} = \frac{- 13.6 z^{2}}{n^{2}} eV E_{2} = \frac{- 13.6}{4} = - 3.4 eV$

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When neutron moving with Kinetic Energy 2eV collides with

stationary $H_{1}^{1}$ in the ground state, the collision will be:

A.

Must be elastic

B.

may be inelastic

C.

Both

D.

Must be perfectly inelastic

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Explanation

If the energy of collision particle is less than that of the first excitation energy of the target atom then the collision must be perfectly elastic.

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Magnetic moment due to the motion of the electron in $n^{th}$ energy state of a hydrogen atom is proportional to

A.

n

B.

$n^{0}$

C.

$n^{5}$

D.

$n^{3}$

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Explanation

$(1) M = (\frac{eh}{4 πm}) n M \propto n$

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The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is:

A.

2

B.

1

C.

4

D.

0.05

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Explanation

(c) Wavelength of spectral lines are given by

$\frac{1}{λ} = z^{2} R (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$

For last line of Balmer series,

$n_{1} = 2 a n d n_{2} = \infty$

$\Rightarrow \frac{1}{λ_{B}} = z^{2} R (\frac{1}{2^{2}} - \frac{1}{\infty^{2}}) = \frac{R}{4} |∴ z = 1|$

similarly,for last line of Lyman series,

$n_{1} = 1 a n d n_{2} = \infty$

$\Rightarrow \frac{1}{λ_{2}} = z^{2} R (\frac{1}{1^{2}} - \frac{1}{\infty^{2}}) = R$

$∴ \frac{\frac{1}{λ_{R}}}{\frac{1}{λ L}} = \frac{4}{R} = \frac{1}{4}$

$\Rightarrow \frac{λ_{L}}{λ_{B}} = \frac{1}{4} \Rightarrow \frac{λ_{B}}{λ_{L}} = 4$

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If an electron in a hydrogen atom jumps from the 3rd orbit, to the 2nd orbit, it emits a photon of wavelength $λ$ . When it jumps from the 4th orbit to the 3rd orbit, the corresponding wavelength of the photon will be

A.

$\frac{16}{25} λ$

B.

$\frac{9}{16} λ$

C.

$\frac{20}{7} λ$

D.

$\frac{20}{13} λ$

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Explanation

(c) Key idea Excess energy of $e^{-}$ appears as photon.

From Rydberg's formula,

$\frac{1}{λ} = R (\frac{1}{n_{1}^{2}} - \frac{1}{n_{1}^{2}}) = R (\frac{1}{2^{2}} - \frac{1}{2^{2}}) = \frac{5 R}{36} \frac{1}{λ} = R (\frac{1}{3^{2}} - \frac{1}{4^{2}}) = \frac{7 R}{144} \frac{1}{λ} / \frac{1}{λ'} = \frac{5 R}{36} + \frac{7 R}{144} \Rightarrow \frac{λ'}{λ} = \frac{5 R}{36} \times \frac{144}{7 R} = \frac{20}{7} \Rightarrow λ' = \frac{20}{7} λ$

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Given the value of Rydberg constant is $10^{7} m^{- 1}$ , the wave number of the last line of the Balmer series in hydrogen spectrum will be:

A.

$0.5 \times 10^{7} m^{- 1}$

B.

$0.25 \times 10^{7} m^{- 1}$

C.

$2.5 \times 10^{7} m^{- 1}$

D.

$0.025 \times 10^{4} m^{- 1}$

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Explanation

(b) Given, Rydberg constant, $R = 10^{7} m^{- 1}$

$∵$ For last time in Balmer series, $n_{2} = \infty, n_{1} = 2 .$

As we know that

$\frac{1}{λ} = R (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) \Rightarrow \frac{1}{λ} = 10^{7} (\frac{1}{2^{2}} - \frac{1}{\infty}) \Rightarrow \bar{V} = \frac{1}{λ} = \frac{10^{7}}{4} = 0.25 \times 10^{7} m^{- 1}$

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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:

A.

$\frac{1}{\sqrt{m}}$

B.

$\frac{1}{m^{2}}$

C.

m

D.

$\frac{1}{m}$

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Explanation

(d) When an $α$ -particle of mass m moving with velocity v bombards on a heavy nucleus of charge Ze, then there will be no loss of energy as in case, initial kinetic enorgy of $α$ -particle potential energy of α-particle at closest approach

=> $\frac{1}{2}$ mv²= $\frac{2 z e^{2}}{4 π ε_{0} r_{0}}$

r_o∝ $\frac{1}{m}$

This is the required closest approach to $α$ -particle from the nucleus

Physics MCQs for NEET — Practice Questions with Answers

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