Question

the momentum of the photon of de-broglie wavelength 6000 A° will be​

Answer 1

Answer:

p = 1.1‬. 10^-27 kgm.m/s

Explanation:

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 6000 Angstrom.

1 A = 10^ -8 cm = 10 ^-10 m

therefore 6000 A = 6. 10^-7 m

therefore

momentum p = h/lambda = 6.63x10 -34 J-s / (6. 10^-7 m)

momentum p = 1.1‬. 10^-27 kgm.m/s

Answer 2

Explanation:

lamda= h/p

6000*10^-10 = 6.626*10^-34/p

p= 6.626*10^-34/6000*10^-10

calculate and get the answer

hope it helps u

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