Q.
Light of wavelength 500 nm is incident on a metal with work function 2.28 eV. The de Broglie wavelength of the emitted electron is
see full answer
Start JEE / NEET / Foundation preparation at rupees 99/day !!
21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!
An Intiative by Sri Chaitanya
a
≥2.8×10-9 m
b
≤2.8×10-12 m
c
<2.8×10-10 m
d
<2.8×10-9 m
answer is A.
(Unlock A.I Detailed Solution for FREE)
Ready to Test Your Skills?
Check your Performance Today with our Free Mock Test used by Toppers!
Take Free Test
Detailed Solution
According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted electron isKmax=hcλ-ϕ0where λ is the wavelength of incident light and ϕ0 is the work function. Here, λ=500nm,hc=1240eVnm and ϕ0=2.28eV∴ Kmax=1240eVnm500nm-2.28eV=2.48eV-2.28eV=0.2eVThe de Broglie wavelength of the emitted electron is λmin=h2mKmaxwhere h is the Planck's constant and m is the mass of the electron. As h=6.6×10-34 J s, m=9×10-31 kg and Kmax=0.2eV=0.2×1.6×10-19 J∴ λmin=6.6×10-34 J s29×10-31 kg0.2×1.6×10-19 J=6.62.4×10-9 m=2.8×10-9 m So, λ≥2.8×10-9 m
Watch 3-min video & get full concept clarity