If h is the planck's constant, the momentum of a photon of wavelength 0.01 A is :

h \times 10^{- 2}

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h \times 10^{2}

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h \times 10^{12}

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Solution

The correct option is D $h \times 10^{12}$
By de-Broglie formula

$λ = \frac{h}{p}$

So, $p = \frac{h}{λ}$

$= \frac{h}{0.01 \times 10^{- 10}} = h \times 10^{12}$

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