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i. The energy of the photon (eV).
ii. The kinetic energy of the emission.
Iii. The velocity of the photoelectron.
Asked by Razaul Last Modified
Answer 
Energy of photon= hc/λ
h=plank constant = 6.63*
h*c= 1240 eV
λ= wavelength of light
E= /
J
E=3.102 eV
2.)Kinetic energy of emission= energy of photon- work function
=3.102-2.13 eV
= 0.972 eV
3.) kinetic energy of electron = 1/2*m* = 0.972 eV
v=
=5.84 m/s
Saikumar A
Tutor
Energy of photon= hc/λ
h=plank constant = 6.63*
c=speed of light=3*
h*c= 1240 eV
λ= wavelength of light
E= /
joule
E=3.102 eV
2.)Kinetic energy of emission= energy of photon- work function
=3.102-2.13 eV
= 0.972 eV
3.)Velocity of free Electrons
kinetic energy of electron = 1/2*m* = 0.972 eV
v=
=5.84 m/s
read less
Durga Prasad Khatri
Tutor
(1) Energy of the photon (E) = hν = hc/λ = (6.626×10-34 Js×3.0×108 ms-1)/
= 4.97×10-19/1.602×10-19 eV
(2) Kinetic energy of emission (1/2 mv2) = hν- hνo = 3.10-2.13 = 0.97 eV
(3) 1/2 mv2 = 0.97 eV = 0.97×1.602×10-19 J
⇒ 1/2×(9.11×10-31 kg)×v2 = 0.97×1.602×10-19 J
⇒ v2 = 0.341×1012 = 34.1×1010
⇒ v = 5.84×105 ms-1
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