In photoelectric effect, the momentum of incident photon of energy $3 \times 10^{- 19} J$ is :

9 \times 10^{11} k g m s^{- 1}

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zero

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10^{- 27} k g m s^{- 1}

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3 \times 10^{- 11} k g m s^{- 1}

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Solution

The correct option is B $10^{- 27} k g m s^{- 1}$
Momentum of photon $P = \frac{E}{c}$
So, $P = \frac{3 \times 10^{- 19}}{3 \times 10^{8}} = 10^{- 27} k g m / s$

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In photoelectric effect, the momentum of incident photon of energy 3×10−19J is :

The correct option is B 10−27kgms−1Momentum 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 :

The correct option is B $10^{- 27} k g m s^{- 1}$
Momentum of photon $P = \frac{E}{c}$
So, $P = \frac{3 \times 10^{- 19}}{3 \times 10^{8}} = 10^{- 27} k g m / s$