Ejection of the photoelectron from metal in t | vedclass

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(C) The energy of the incident radiation is given by $E = \frac{hc}{\lambda}$.Using $hc = 12400\, eV\, \mathring{A}$ and $\lambda = 250\, nm = 2500\,

Vedclass · Accepted Answer

$4.5$

Vedclass · Answer

(C) The energy of the incident radiation is given by $E = \frac{hc}{\lambda}$.Using $hc = 12400\, eV\, \mathring{A}$ and $\lambda = 250\, nm = 2500\, \mathring{A}$,we get $E = \frac{12400}{2500} = 4.96\, eV$.According to the photoelectric equation,$E = W_0 + K.E._{max}$,where $W_0$ is the work function and $K.E._{max}$ is the maximum kinetic energy.The stopping potential is $0.5\, V$,so $K.E._{max} = 0.5\, eV$.Substituting the values: $4.96 = W_0 + 0.5$.Therefore,$W_0 = 4.96 - 0.5 = 4.46\, eV$,which is approximately $4.5\, eV$.

List-$I$ (Spectral Series for Hydrogen)	List-$II$ (Spectral Region)
$A$. Lyman	$I$. Infrared region
$B$. Balmer	$II$. $UV$ region
$C$. Paschen	$III$. Infrared region
$D$. Pfund	$IV$. Visible region

Ejection of the photoelectron from metal in the photoelectric effect experiment can be stopped by applying $0.5\, V$ when the radiation of $250\, nm$ is used. The work function of the metal is ................ $eV$

Similar Questions

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The energy required (in $eV$) to excite an electron of $H$-atom from the ground state to the third excited state is

The series limit for the Balmer series of $H$ atom spectra is:

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Match List-$I$ with List-$II$.

List-$I$ (Spectral Series for Hydrogen) List-$II$ (Spectral Region)

$A$. Lyman $I$. Infrared region

$B$. Balmer $II$. $UV$ region

$C$. Paschen $III$. Infrared region

$D$. Pfund $IV$. Visible region

Choose the correct answer from the options given below:

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