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Q.

There are $n_{1}$ photons of frequency $v_{1}$ in a beam of light. In an equally energetic beam, there are $n_{2}$ photons of frequency $v_{2}$ . Then the correct relation is

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a

$\frac{n_{1}}{n_{2}} = 1$

b

$\frac{n_{1}}{n_{2}} = \frac{v_{1}}{v_{2}}$

c

$\frac{n_{1}}{n_{2}} = \frac{v_{2}}{v_{1}}$

d

$\frac{n_{1}}{n_{2}} = \frac{v_{1}^{2}}{v_{2}^{2}}$

answer is C.

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Detailed Solution

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$E = nhv \Rightarrow v \propto \frac{1}{n} \Rightarrow \frac{n_{1}}{n_{2}} = \frac{v_{2}}{v_{1}} .$

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There are n1 photons of frequency v1 in a beam of light. In an equally energetic beam, there are n2 photons of frequency v2. Then the correct relation is