- Home
- Standard 12
- Physics
There are two sources of light, each emitting with a power of $100 \,W.$ One emits $X-$ rays of wavelength $1\, nm$ and the other visible light at $500\, nm$. Find the ratio of number of photons of $X-$ rays to the photons of visible light of the given wavelength ?
Solution
Power of radiation,
$\mathrm{P}=\frac{\mathrm{E}_{n}}{t}=\frac{n h f}{t}=\frac{n h c}{t \lambda}$
$\therefore \mathrm{P}=n^{\prime} \frac{h c}{\lambda} \quad$ (Where $n^{\prime}=$ no. of photons emitted per unit time)
$\therefore n^{\prime}=\left(\frac{\mathrm{P}}{h c}\right) \lambda$ $\therefore n^{\prime} \propto \lambda \quad(\because$ Here P, $h, c$ are constant $)$ $\therefore \frac{n_{1}^{\prime}}{n_{2}^{\prime}}=\frac{\lambda_{1}}{\lambda_{2}}$ $=\frac{1 \mathrm{~nm}}{500 \mathrm{~nm}}$ $\therefore \frac{n_{1}^{\prime}}{n_{2}^{\prime}}=\frac{1}{500}$
$\therefore\frac{n_{1}^{\prime}}{n_{2}^{\prime}}=\frac{\lambda_{1}}{\lambda_{2}}$
$=\frac{1 \mathrm{~nm}}{500 \mathrm{~nm}}$
$\therefore \frac{n_{1}^{\prime}}{n_{2}^{\prime}}=\frac{1}{500}$
