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JEE Main 20209 JanuaryEvening Shift

Exam: JEE Main
Year: 2020
Date: 9 January
Shift: Evening Shift
Subject: Physics
Chapter: Atomic Physics

JEE Main 2020 · 9 January · Evening Shift · Physics · Atomic Physics

The energy required to ionise a hydrogen like ion in its ground...

The energy required to ionise a hydrogen like ion in its ground state is

9

Rydbergs. What is the wavelength of the radiation emitted when the electron in this ion jumps from the second excited stale to the ground state?

Options

A $24 . 2nm$
B $11.4nm$
C $35.8nm$
D $8.6nm$

Correct answer

B. $11.4nm$

Step-by-step solution

\dfrac{hc}{\lambda} = \left(13 . 6eV\right){z}^{2}\left[\dfrac{1}{{n}_{1}^{2}} - \dfrac{1}{{n}_{2}^{2}}\right]

{n}_{1} = 1

{n}_{2} = 3

\dfrac{hc}{\lambda} = \left(13.6eV\right)\left({3}^{2}\right)\left[\dfrac{1}{{1}^{2}} - \dfrac{1}{{3}^{2}}\right]

⇒ \dfrac{hc}{\lambda} = \left(13.6eV\right)\left(9\right)\left(\dfrac{8}{9}\right)

Wavelength

= \dfrac{1240}{8 \times 13.6}\text{nm}

\lambda = 11.39\text{nm}

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