QJEEAdv18Ph0642: JEE Advanced 2018 Physics - Class 12 JEEAdv18Ph0642 Solution: Consider a hydrogen-like ionized atom with atomic number with a …

Question

Verified Solution • 2025–26 Session moderate

Consider a hydrogen-like ionized atom with atomic number

Z

with a single electron. In the emission spectrum of this atom, the photon emitted in the

n=2

to

n=1

transition has energy

74.8eV

higher than the photon emitted in the

n=3

to

n=2

transition. The ionization energy of the hydrogen atom is

13.6

eV.

The value of

Z

is ____________.

Accepted Answer

We know that,

$\Delta {E_{i - j}} = 13.6\left[ {{1 \over {n_j^2}} - {1 \over {n_i^2}}} \right]{Z^2}$

Therefore,

$\Delta {E_{2 - 1}} = 13.6\left[ {{1 \over {{1^2}}} - {1 \over {{2^2}}}} \right]{Z^2} = 13.6\left[ {{1 \over 1} - {1 \over 4}} \right]{Z^2}$

$\Delta {E_{2 - 1}} = 13.6 \times {3 \over 4} \times {Z^2}$

Similarly,

$\Delta {E_{3 - 2}} = 13.6\left[ {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right]{Z^2} = 13.6\left[ {{1 \over 4} - {1 \over 9}} \right]{Z^2}$

$\Delta {E_{3 - 2}} = 13.6 \times {5 \over {36}} \times {Z^2}$

Given, $\Delta {E_{2 - 1}} = \Delta {E_{3 - 2}} + 74.8$

$\Rightarrow 13.6 \times {3 \over 4} \times {Z^2} = 13.6 \times {5 \over {36}} \times {Z^2} + 74.8$

$\Rightarrow 13.6 \times {3 \over 4} \times {Z^2} - 13.6 \times {5 \over {36}} \times {Z^2} = 74.8$

$\Rightarrow 13.6 \times {Z^2}\left[ {{3 \over 4} - {5 \over {36}}} \right] = 74.8$

$\Rightarrow 13.6 \times {Z^2}\left[ {{{27 - 5} \over {36}}} \right] = 74.8$

$\Rightarrow 13.6 \times {Z^2} \times {{22} \over {36}} = 74.8$

$\Rightarrow {Z^2} = {{74.8 \times 36} \over {13.6 \times 22}} = 9$

$\Rightarrow Z = 3$

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