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NEET NEET Questions — Free MCQ Practice

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The shortest wavelength of Balmer series is about

A.

912 $\overset{o}{A}$

B.

121 $\overset{o}{A}$

C.

3648 $\overset{o}{A}$

D.

4864 $\overset{o}{A}$

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Explanation

$(3) \frac{1}{λ} = R \{\frac{1}{4} - \frac{1}{\infty}\} = 3648 \overset{o}{A}$

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$If λ_{1} be longest wavelength of Lyman series and λ_{2} be the longest wavelength of Balmer series for a hydrogen atom . What is the wavelength of photon emitted when an electron makes a transition from n = 3 to n = 1 in a hydrogen atom ?$

A.

$λ_{1} + λ_{2}$

B.

${(\frac{1}{λ_{1}} + \frac{1}{λ_{2}})}^{- 1}$

C.

$\frac{λ_{1} + λ_{2}}{2}$

D.

$\sqrt{λ_{1} λ_{2}}$

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Explanation

$(2) L y m a n s e r i e s : \frac{1}{λ_{1}} = R [\frac{1}{1^{2}} - \frac{1}{2^{2}}] E n e r g y o f p h o t o n E_{1} = \frac{h c}{λ_{1}} B a l m e r s e r i e s : \frac{1}{λ_{2}} = R [\frac{1}{2^{2}} - \frac{1}{3^{2}}] E n e r g y o f p h o t o n E_{2} = \frac{h c}{λ_{2}} F o r t r a n s i t i o n f r o m n = 3 t o n = 1, E = E_{3 \to 2} + E_{2 \to 1} = E_{2} + E_{1} \frac{hc}{λ_{3}} = \frac{hc}{λ_{1}} + \frac{hc}{λ_{2}}$

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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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The potential difference applied to an X-ray tube is 5 kV and current through it is 3.2 mA. The number of electrons striking the target per sec is:

A.

$2 \times 10^{16}$

B.

$5 \times 10^{6}$

C.

$1 \times 10^{17}$

D.

$4 \times 10^{15}$

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Explanation

$(1) I = ne n = \frac{3.2 x 10^{- 3}}{1.6 x 10^{- 19}} = 2 x 10^{16}$

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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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