Photoelectric EffectEasy
Question
The maximum kinetic energy of photo-electron liberated from the surface of lithium
(work function = 2.35 eV) by electromagnetic radiation whose electric component varies with time as
E = a[1 + cos(2πf1t)] cos(2πf2t) where ′a′ is a constant, f1 = 3.6 × 1015 Hz and f2 = 1.2 × 1015 Hz is
[Take : h = 6.6 × 10-34 J-s]
(work function = 2.35 eV) by electromagnetic radiation whose electric component varies with time as
E = a[1 + cos(2πf1t)] cos(2πf2t) where ′a′ is a constant, f1 = 3.6 × 1015 Hz and f2 = 1.2 × 1015 Hz is
[Take : h = 6.6 × 10-34 J-s]
Options
A.2. 6 eV
B.7.55 eV
C.12.5 eV
D.17.45 eV
Solution
E = a(1 + cos2πf1t)cos2πf2t = a cos2πf2t + a cos2πf1t cos2πf2t
⇒ E = a cos2πf2t +
acos2π(f1 + f2)t + 
acos2π(f1 - f2)t
This is a complex vibration consisting of harmonic vibrations of frequencies f2, (f1 + f2)and (f1 - f2)
The highest is (f1 + f2). So, hv = φ + Tmax
Tmax = h(f1 + f2) - φ =
( 3.6 × 1015 + 1.2 × 1015 ) - 2.35 = 17.45 eV
⇒ E = a cos2πf2t +
acos2π(f1 + f2)t + acos2π(f1 - f2)tThis is a complex vibration consisting of harmonic vibrations of frequencies f2, (f1 + f2)and (f1 - f2)
The highest is (f1 + f2). So, hv = φ + Tmax
Tmax = h(f1 + f2) - φ =
( 3.6 × 1015 + 1.2 × 1015 ) - 2.35 = 17.45 eVCreate a free account to view solution
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