momentum of the photon of wavelength 5000A will be??
Answers
Answered by
3
Answer:
1.3×10−27kg−m/sec
1.3×10−28kg−m/sec
4×1029kg−m/sec
4×10−18kg−m/sec
Answer :
A
Solution :
p=hλ=6.6×10−34(5000×10−10)=1.3×10−27kg−m/s
Answered by
3
using de Broglie relation for photon
wavelength = h/p
where h is Planck's constant h= 6.63x10 -34 J-s and p is momentum.
wavelength(lambda) is given as 5000 Angstrom.
1 A = 10^ -8 cm = 10 ^-10 m
therefore 5000 A = 5. 10^-7 m
therefore
momentum p = h/lambda = 6.63x10 -34 J-s / (5. 10^-7 m)
momentum p = 1.32. 10^-27 kgm.m/s
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