If $n_{r}$ and $n_{b}$ be respectively the number of photons emitted by a red bulb and a blue bulb of equal power in one second, then $n_{r}$ is $n_{b} .$

equal to

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less than

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greater than

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Solution

The correct option is C greater than
Let $v_{r}$ and $v_{b}$ be the freqencies of red and blue light respectively. We know that $v_{b} > v_{r}$ .

Now, since the two sources emit with equal power, the energy emitted per second will be -

$n_{r} (h v_{r}) = n_{b} (h v_{b}),$

which would imply,

$n_{r} v_{r} = n_{b} v_{b} .$

Since $v_{b} > v_{r}$ , we can directly conclude: $n_{r} > n_{b}$ .

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