JEE Advanced 2019 Paper 1 · Physics · Atoms, Nuclei and Dual Nature of Radiation · hard

A photon emitted in the n=3→ n=2 transition of a hydrogen-like Li^2+ ion falls on a metal surface of work function 4.0 eV. Use En=-13.6 Z^2/n^2 eV…

Single correct option

A photon emitted in the

n=3\to n=2

transition of a hydrogen-like

\mathrm{Li^{2+}}

ion falls on a metal surface of work function

4.0\ \mathrm{eV}

. Use

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

and

\lambda(\mathrm{nm})=1.226/\sqrt{K(\mathrm{eV})}

for a free electron of kinetic energy

K

. The de Broglie wavelength of the fastest photoelectron is closest to

A. $0.30\ \mathrm{nm}$
B. $0.34\ \mathrm{nm}$
C. $0.61\ \mathrm{nm}$
D. $0.17\ \mathrm{nm}$

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Answer: (B) $0.34\ \mathrm{nm}$

For

\mathrm{Li^{2+}}

Z=3

. The photon energy is

E=13.6\times9\left(\dfrac14-\dfrac19\right)=13.6\times9\times\dfrac{5}{36}=17.0\ \mathrm{eV}

. By the photoelectric equation the maximum kinetic energy is

K_{max}=17.0-4.0=13.0\ \mathrm{eV}

. For the fastest electron the de Broglie wavelength is

\lambda=h/\sqrt{2mK}=1.226/\sqrt{13}\ \mathrm{nm}=0.340\ \mathrm{nm}

. Using the photon energy itself (17 eV) gives

0.297\ \mathrm{nm}

, i.e.

0.30\ \mathrm{nm}

, and using

K=4\ \mathrm{eV}

gives

0.61\ \mathrm{nm}

. Forgetting the

Z^2

factor gives a photon of

1.89\ \mathrm{eV}

, too low to eject any electron from this metal. Each distractor therefore comes from a plausible slip in the photon energy, the work-function subtraction or the charge number. The answer is option (B).

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