JEE Main 2018 · Physics · Atoms, Nuclei and Dual Nature of Radiation · easy

A singly ionised helium ion He^+ makes a transition from the n=4 level to the n=2 level. Using the Bohr model with En=-13.6 Z^2/n^2 eV and hc=1240 eV…

Single correct option

A singly ionised helium ion

\mathrm{He^+}

makes a transition from the

n=4

level to the

n=2

level. Using the Bohr model with

E_n=-13.6\,Z^2/n^2\ \mathrm{eV}

and

hc=1240\ \mathrm{eV\,nm}

, the wavelength of the emitted photon is nearly

Show answer and solution

Answer: (D) $121.6\ \mathrm{nm}$

For

\mathrm{He^+}

Z=2

. The photon energy is

\Delta E=13.6\times4\left(\dfrac1{2^2}-\dfrac1{4^2}\right)=13.6\times4\times\dfrac{3}{16}=10.2\ \mathrm{eV}

. The wavelength is

\lambda=\dfrac{1240}{10.2}=121.6\ \mathrm{nm}

, which is the same as the Lyman-

\alpha

line of hydrogen (

n=2\to1

10.2\ \mathrm{eV}

). The values

486

and

656\ \mathrm{nm}

are hydrogen Balmer lines, and

30.4\ \mathrm{nm}

is the

n=2\to1

line of

\mathrm{He^+}

Original question written in the JEE Main 2018 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake