First slide

Matter waves (DC Broglie waves)

Question

The momentum of a photon of energy 1 MeV in kg m/s will be

Easy

A

$0.33 \times 10^{6}$
B

$7 \times 10^{- 24}$
C

$10^{- 22}$
D

$5 \times 10^{- 22}$

Solution

Energy of photon is given by
$\begin{array}{l} E = \frac{hc}{λ} . . . (i) \\ Where h is plank' s constant, c the velocity of light and λ its wavelength . \\ de - Broglie wavelength is given by \\ λ = \frac{h}{p} . . . (ii) \\ P being momentum of photon from Eq . s (i) and (ii), we can have \\ E = \frac{hc}{h / p} = pc or p = E / c \\ Given, E = 1 MeV = 1 \times 10^{6} \times 1 .6 \times 10^{- 19} J \\ c = 3 \times 10^{8} m / s \\ Hence, after putting numerical values, we obtain p = \frac{1 \times 10^{6} \times 1 .6 \times 10^{- 19}}{3 \times 10^{8}} kgm / s \\ = 5 \times 10^{- 22} kgm / s \end{array}$

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