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Q.

Energy of photon whose frequency is $10^{12}$ MHz, will be

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a

$4 .14 \times 10^{3} eV$

b

$4 .14 \times 10^{3} keV$

c

$4 .14 \times 10^{2} keV$

d

$4 .14 \times 10^{3} MeV$

answer is D.

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Detailed Solution

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$E (eV) = \frac{hv}{e} = \frac{6 .0 \times 10^{- 34} \times 10^{12} \times 10^{6}}{1 .6 \times 10^{- 19}} = 4 .14 \times 10^{3} eV$

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