The electron of a Li^2+ ion undergoes…

The electron of a Li^2+ ion undergoes a transition from n=3 to n=2 and the emitted photon falls on a metal surface of work function 4.5 eV. The…

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The electron of a

\mathrm{Li^{2+}}

ion undergoes a transition from

n=3

n=2

and the emitted photon falls on a metal surface of work function

4.5\ \mathrm{eV}

. The photoelectrons emitted with maximum kinetic energy enter, perpendicular to the field, a region of uniform magnetic field

B=1.0\ \mathrm{mT}

. Take the ground-state energy of hydrogen as

-13.6\ \mathrm{eV}

e=1.6\times10^{-19}\ \mathrm{C}

and

m_e=9.11\times10^{-31}\ \mathrm{kg}

. The radius of the circular path of these electrons is closest to

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Answer: (A) $11.9\ \mathrm{mm}$

For a hydrogen-like ion

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

. With

Z=3

h\nu=13.6\times9\left(\tfrac14-\tfrac19\right)=13.6\times9\times\tfrac5{36}=17.0\ \mathrm{eV}

. Photoelectric equation:

K_{\max}=17.0-4.5=12.5\ \mathrm{eV}

. In the magnetic field,

r=\dfrac{p}{eB}=\dfrac{\sqrt{2m_eK}}{eB}

with

K=12.5\times1.6\times10^{-19}\ \mathrm{J}

p=\sqrt{2\times9.11\times10^{-31}\times2.0\times10^{-18}}=1.909\times10^{-24}\ \mathrm{kg\,m\,s^{-1}}

, so

r=\dfrac{1.909\times10^{-24}}{1.6\times10^{-19}\times10^{-3}}=1.19\times10^{-2}\ \mathrm{m}\approx11.9\ \mathrm{mm}

. Forgetting the work function (

K=17

eV) gives

13.9

mm; the other values come from dropping the square root factor of

2

or doubling

r

Original question written in the JEE Advanced 2025 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake