JEE Main 2025 · Physics · Atoms, Nuclei and Dual Nature of Radiation · medium

A photon emitted when a He^+ ion makes a transition from the n=4 level to the n=2 level falls on a metal surface of work function 4.2 eV. Taking the…

Single correct option

A photon emitted when a

\mathrm{He^+}

ion makes a transition from the

n=4

level to the

n=2

level falls on a metal surface of work function

4.2\ \mathrm{eV}

. Taking the hydrogen ground-state energy as

-13.6\ \mathrm{eV}

, the stopping potential for the emitted photoelectrons is

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Answer: (B) $6.0\ \mathrm{V}$

For a hydrogen-like ion

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

. For

\mathrm{He^+}

(

Z=2

) the photon energy is

h\nu = 13.6\times4\left(\dfrac14-\dfrac1{16}\right) = 54.4\times\dfrac{3}{16} = 10.2\ \mathrm{eV}

. Einstein's photoelectric equation gives

K_{\max} = h\nu-\phi = 10.2-4.2 = 6.0\ \mathrm{eV}

. The stopping potential satisfies

eV_s = K_{\max}

, so

V_s = 6.0\ \mathrm{V}

. The value

10.2\ \mathrm{V}

ignores the work function.

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