Let n_r and n_b be respectively the number of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$n_r > n_b$

Vedclass · Answer

(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.

Similar Questions

$A$ blue lamp emits light of mean wavelength $4500 \ \text{Å}$. The lamp is rated at $150 \ \text{W}$ and $8 \%$ efficiency. Calculate the number of photons emitted by the lamp per second.

According to the modern theory of the nature of light,light has ...

The momentum of a photon is $2 \times 10^{-16} \text{ g} \cdot \text{cm/s}$. Its energy is:

The momentum of a photon with energy $20 \, eV$ is

The number of photons emitted per second from a $60 \ W$ light bulb is .......... . The wavelength of the photon is $660 \ nm$. $(h = 6.6 \times 10^{-34} \ J \cdot s)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module