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Physics MCQs for NEET — Practice Questions with Answers

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If the binding energy of the electron in a hydrogen atom is 13.6 eV, the energy required to remove the electron from the first excited state of ${Li}^{+ +}$ is

122.4 eV

30.6 eV

13.6 eV

3.4 eV

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Explanation

(b) $E_{n} = \frac{13.6}{n^{2}} \times Z^{2}$

For first excited state n = 2 and for ${Li}^{+ +}$ , z =3 $\Rightarrow E = \frac{13.6}{4} \times 9 = 30.6 eV$

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Which state of triply ionised Beryllium $({Be}^{+ + +})$ has the same orbital radius as that of the ground state of hydrogen

n = 4

n = 3

n = 2

n = 1

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Explanation

(c) Radius of nth orbit for any hydrogen like atom
$r_{n} = r_{0} (\frac{n^{2}}{Z})$ ( $r_{0}$ = radius of first orbit of $H_{2}$ -atom)
If $r_{n} = r_{0} \Rightarrow n = \sqrt{Z}$ . For ${Be}^{+ + +}$ , Z = 4 $\Rightarrow$ n = 2

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Taking Rydberg’s constant $R_{H} = 1.097 \times 10^{7} m$ first and second wavelength of Balmer series in hydrogen spectrum is

2000 Å, 3000 Å

1575 Å, 2960 Å

6529 Å, 4280 Å

6552 Å, 4863 Å

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Explanation

(d) $\frac{1}{λ} = R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$ For first wavelength, $n_{1} = 2$ , $n_{2} = 3$ $\Rightarrow λ_{1} = 6563$ Å. For second wavelength, $n_{1} = 2, n_{2} = 4 \Rightarrow λ_{2} = 4861$ Å

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The kinetic energy of electron in the first Bohr orbit of the hydrogen atom is

– 6.5 eV

– 27.2 eV

13.6 eV

– 13.6 eV

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Explanation

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In Bohr’s model of hydrogen atom, which of the following pairs of quantities are quantized

Energy and linear momentum

Linear and angular momentum

Energy and angular momentum

None of the above

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Explanation

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The energy of the highest energy photon of Balmer series of hydrogen spectrum is close to

13.6 eV

3.4 eV

1.5 eV

0.85 eV

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Explanation

(b) $E = 13.6 [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$ For highest energy in Balmer series $n_{1} = 2$ and $n_{2} = \infty$

$\Rightarrow E = 13.6 [\frac{1}{2^{2}} - \frac{1}{\infty^{2}}] = 3.4 eV$

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Which one of the relation is correct between time period and number of orbits while an electron is revolving in a orbit

$n^{2}$

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

$n^{3}$

$\frac{1}{n}$

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Explanation

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If the energy of a hydrogen atom in nth orbit is $E_{n}$ , then energy in the nth orbit of a singly ionized helium atom will be

4 $E_{n}$

$E_{n}$ /4

2 $E_{n}$

$E_{n}$ /2

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Explanation

(a) $E_{n} \propto Z^{2} \Rightarrow \frac{{(E_{n})}_{He}}{{(E_{n})}_{H}} = \frac{Z_{He}^{2}}{Z_{H}^{2}} = 4 \Rightarrow {(E_{n})}_{He} = 4 \times {(E_{n})}_{H}$

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What is the ratio of wavelength of radiations emitted when an electron in hydrogen atom jump from fourth orbit to second orbit and from third orbit to second orbit

27 : 25

20 : 27

20 : 25

25 : 27

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Explanation

(b) By using $\frac{1}{λ} = R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$

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As the electron in Bohr orbit of Hydrogen atom passes from state n = 2 to n = 1 , the kinetic energy K and potential energy U change as

K two-fold, U four-fold

K four-fold, U two-fold

K four-fold, U also four-fold

K two-fold, U also two-fold

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Explanation

(c)

$K E \propto \frac{1}{n^{2}} \Rightarrow \frac{K_{1}}{K_{2}} = \frac{{n_{2}}^{2}}{{n_{1}}^{2}} = \frac{4}{1} P E \propto \frac{1}{n^{2}} \Rightarrow \frac{U_{1}}{U_{2}} = \frac{{n_{2}}^{2}}{{n_{1}}^{2}} = \frac{4}{1}$

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