If n_r and n_b are the number of photons emit | vedclass

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(C) The power $P$ of a light source is given by $P = n \cdot E$,where $n$ is the number of photons emitted per second and $E = \frac{hc}{\lambda}$ is

Vedclass · Accepted Answer

$n_r > n_b$

Vedclass · Answer

(C) The power $P$ of a light source is given by $P = n \cdot E$,where $n$ is the number of photons emitted per second and $E = \frac{hc}{\lambda}$ is the energy of a single photon.Since the power $P$ is the same for both bulbs,we have $n_r E_r = n_b E_b$.This implies $n_r \left( \frac{hc}{\lambda_r} \right) = n_b \left( \frac{hc}{\lambda_b} \right)$,which simplifies to $\frac{n_r}{\lambda_r} = \frac{n_b}{\lambda_b}$.Since the wavelength of red light is greater than that of blue light $(\lambda_r > \lambda_b)$,it follows that $n_r > n_b$ to maintain the equality of power.Therefore,the number of photons emitted by the red bulb is greater than that of the blue bulb.

If $n_r$ and $n_b$ are the number of photons emitted per second by a red bulb and a blue bulb of equal power respectively,then:

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