The ratio of the energy of a photon with lamb | vedclass

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

Question

(A) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$,where $h$ is Planck's constant,$c$ is the speed of light,and $\lambda$ is

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$,where $h$ is Planck's constant,$c$ is the speed of light,and $\lambda$ is the wavelength.Since $h$ and $c$ are constants,the energy $E$ is inversely proportional to the wavelength $\lambda$,i.e.,$E \propto \frac{1}{\lambda}$.Let $E_1$ be the energy of the photon with $\lambda_1 = 150\,nm$ and $E_2$ be the energy of the photon with $\lambda_2 = 300\,nm$.Then,the ratio is $\frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1}$.Substituting the given values: $\frac{E_1}{E_2} = \frac{300\,nm}{150\,nm} = 2$.Therefore,the ratio is $2:1$.

The ratio of the energy of a photon with $\lambda = 150\,nm$ to that with $\lambda = 300\,nm$ is

Similar Questions

$A$ photon of frequency $f$ has a momentum associated with it. If $c$ is the velocity of light,the momentum is:

$1 \ J$ is equal to how many electron volt?

Monochromatic light of wavelength $3000 \, \mathring{A}$ is incident on a surface of area $4 \, cm^2$. If the intensity of light is $150 \, mW/m^2$, then the rate at which photons strike the target is:

$A$ stick of $1 \, m$ is moving with a velocity of $2.7 \times 10^8 \, m/s$. What is the apparent length of the stick in meters? (Given: $c = 3 \times 10^8 \, m/s$)

$A$ $10\, kW$ transmitter emits radio waves of wavelength $500\, m$. The number of photons emitted per second by the transmitter is of the order of

Vedclass Test Series

Exam Paper Generator

Online Exam Module