In photoelectric effect- the momentum of incident photon of energy 3 times 10-19- J is - zero10-27- kgms-1-9 times 10-11- kgms-1-3 times 10-11- kgms-1-

Question

Toppr Admin · Accepted Answer

Momentum of photon-xA0- -xA0- -xA0-P-EcSo-xA0- -xA0- P-3-xD7-10-x2212-193-xD7-108-10-x2212-27-xA0-kg-xA0-m-s

In photoelectric effect, the momentum of incident photon of energy $3 \times 10^{- 19} J$ is :
zero
$10^{- 27} k g m s^{- 1}$
$9 \times 10^{11} k g m s^{- 1}$
$3 \times 10^{- 11} k g m s^{- 1}$

Momentum of photon $P = \frac{E}{c}$
So, $P = \frac{3 \times 10^{- 19}}{3 \times 10^{8}} = 10^{- 27} k g m / s$

In photoelectric effect, the momentum of incident photon of energy 3×10−19J is : zero10−27kgms−19×1011kgms−13×10−11kgms−1

Momentum of photon P=EcSo, P=3×10−193×108=10−27 kg m/s

In photoelectric effect, the momentum of incident photon of energy $3 \times 10^{- 19} J$ is :
zero
$10^{- 27} k g m s^{- 1}$
$9 \times 10^{11} k g m s^{- 1}$
$3 \times 10^{- 11} k g m s^{- 1}$

Momentum of photon $P = \frac{E}{c}$
So, $P = \frac{3 \times 10^{- 19}}{3 \times 10^{8}} = 10^{- 27} k g m / s$