Two monochromatic beams $A$ and $B$ of equal intensity $I$, hit a screen. The number of photons hitting the screen by beam $A $ is twice that by beam $ B$. Then what inference can you make about their frequencies ?
Intensity of light is given by,
$\mathrm{I}=\frac{\mathrm{E}_{n}}{\mathrm{~A} t}=\frac{n h f}{\mathrm{~A} t}$
(Where $A=$ area of given screen)
$\therefore \mathrm{I}=n^{\prime} hf$
where $n^{\prime}=\frac{n}{\mathrm{~A} t}=$ no. of photons incident on unit area of screen per unit time
$\therefore n^{\prime} f=\frac{\mathrm{I}}{h}=$ constant $(\because$ Here $\mathrm{I}$ is constant $)$
$\therefore n_{\mathrm{A}}^{\prime} f_{\mathrm{A}}=n_{\mathrm{B}}^{\prime} f_{\mathrm{B}}$
$\therefore\left(2 n_{\mathrm{B}}^{\prime}\right) f_{\mathrm{A}}=n_{\mathrm{B}}^{\prime} f_{\mathrm{B}} \quad$ (As per the statement)
$\therefore f_{\mathrm{A}}=\frac{f_{\mathrm{B}}}{2}$
$\Rightarrow$ Frequency of beam $A$ would be half of frequency of beam $B$.
A photon collides with a stationary hydrogen atom in ground state inelastically. Energy of the colliding photon is $10.2 \ eV$. After a time interval of the order of micro second another photon collides with same hydrogen atom inelastically with an energy of $15 \ eV$. What will be observed by the detector
The momentum of a photon in an $X-$ ray beam of ${10^{ - 10}}$ metre wavelength is
Using the Heisenberg uncertainty principle, arrange the following particles in the order of increasing lowest energy possible.
$(I)$ An electron in $H _{2}$ molecule
$(II)$ A hydrogen atom in a $H _{2}$ molecule
$(III)$ A proton in the carbon nucleus
$(IV)$ $A H _{2}$ molecule within a nanotube
In a photoelectric effect experiment a light of frequency $1.5$ times the threshold frequency is made to fall on the surface of photosensitive material. Now if the frequency is halved and intensity is doubled, the number of photo electrons emitted will be: