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 74.8 eV 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 ____.

JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Modern Physics

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

n = 1

transition has energy 74.8 eV higher than the photon emitted in the

n = 3

n = 2

transition. The ionization energy of the hydrogen atom is

13.6 e V

. The value of

Z

is ____.

Answer: 3

The energy of level

n

E_{n} = - 13.6 \frac{Z ^{2}}{n ^{2}}

eV. The energy of the

2 \to 1

transition is

E_{2 \to 1} = 13.6 Z^{2} (1 - \frac{1}{4}) = 13.6 Z^{2} \times \frac{3}{4}

. The energy of the

3 \to 2

transition is

E_{3 \to 2} = 13.6 Z^{2} (\frac{1}{4} - \frac{1}{9}) = 13.6 Z^{2} \times \frac{5}{36}

. The difference is

13.6 Z^{2} (\frac{3}{4} - \frac{5}{36}) = 13.6 Z^{2} \times \frac{22}{36} = 74.8

, giving

Z^{2} = 9

, so

Z = 3

JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Numerical answer