When light of wavelength 248 nm falls on a metal of threshold energy 3.0 eV , the de-Broglie wavelength of emitted electrons is \_\_\_\_ \(\stackrel{\circ}{\text{A}}\). (Round off to the nearest Integer). [Use : \(\sqrt{3}=1.73, \mathrm{~h}=6.63 \times 10^{-34} \mathrm{Js}, \mathrm{m}_{\mathrm{e}}=9.1 \times 10^{-31} \mathrm{~kg}, \mathrm{c}=3.0 \times 10^{8} \mathrm{~ms}^{-1}, 1\mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}\)]

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Atomic Structure

When light of wavelength 248 nm falls on a metal of threshold energy 3.0 eV , the de-Broglie wavelength of emitted electrons is ____

A \circ

.
(Round off to the nearest Integer). [Use :

3 = 1.73, h = 6.63 \times 1 0^{- 34} Js, m_{e} = 9.1 \times 1 0^{- 31} kg, c = 3.0 \times 1 0^{8} ms^{- 1}, 1 eV = 1.6 \times 1 0^{- 19} J

]

Answer: 9

The incident photon carries

E = \frac{h c}{λ} = \frac{6.63 \times 1 0 ^{- 34} \times 3.0 \times 1 0 ^{8}}{248 \times 1 0 ^{- 9}} = 8.02 \times 1 0^{- 19} J = 5.0 eV

Of this, 3.0 eV is spent in freeing the electron, so the electron leaves with

K = 5.0 - 3.0 = 2.0 eV = 3.2 \times 1 0^{- 19} J

Its momentum is

p = 2 m_{e} K = 2 \times 9.1 \times 1 0^{- 31} \times 3.2 \times 1 0^{- 19} = 7.63 \times 1 0^{- 25} kg m s^{- 1}

, so

λ_{dB} = \frac{h}{p} = \frac{6.63 \times 1 0 ^{- 34}}{7.63 \times 1 0 ^{- 25}} = 8.7 \times 1 0^{- 10} m

Rounded to the nearest integer, the de Broglie wavelength of the emitted electrons is 9 Å.

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Integer answer