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(C) The power $P$ of a bulb is given by the total energy emitted per unit time,which is $P = n \cdot E_{photon} = n \cdot \frac{hc}{\lambda}$.Since bo

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$n_r > n_b$

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(C) The power $P$ of a bulb is given by the total energy emitted per unit time,which is $P = n \cdot E_{photon} = n \cdot \frac{hc}{\lambda}$.Since both bulbs have equal power $P$ and operate for the same time,the total energy emitted is the same.Therefore,$n_r \cdot \frac{hc}{\lambda_r} = n_b \cdot \frac{hc}{\lambda_b}$.This simplifies to $n_r \cdot \frac{1}{\lambda_r} = n_b \cdot \frac{1}{\lambda_b}$,or $n_r = n_b \cdot \frac{\lambda_r}{\lambda_b}$.We know that the wavelength of red light is greater than the wavelength of blue light,i.e.,$\lambda_r > \lambda_b$.Since $\frac{\lambda_r}{\lambda_b} > 1$,it follows that $n_r > n_b$.

Let $n_r$ and $n_b$ be respectively the number of photons emitted by a red bulb and a blue bulb of equal power in a given time.

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