The graph shows variation of stopping potential with the frequency of the incident radiation for three photosensitive metals

Question

The graph shows variation of stopping potential $V_{0}$ with the frequency $ν$ of the incident radiation for three photosensitive metals $X_{1}$ , $X_{2}$ and $X_{3}$ . Which metal will give out electrons with greater kinetic energy, for the same wavelength of incident radiation [2026]

Smart Achievers · Accepted Answer

(1)

$V_{0} = \frac{E}{e} - \frac{ϕ}{e}$

$X_{3}$

$K E_{\max} is highest for least ν_{th}$

$As metal (1) having minimum value of work function$

$So X_{1} will have maximum kinetic energy$

Solution

Q.
The graph shows variation of stopping potential $V_{0}$ with the frequency $ν$ of the incident radiation for three photosensitive metals $X_{1}$ , $X_{2}$ and $X_{3}$ . Which metal will give out electrons with greater kinetic energy, for the same wavelength of incident radiation [2026]

1 $X_{1}$

2 $X_{2}$

3 $X_{3}$

4 All the metals will give out photo electrons with same kinetic energies.

Ans.
(1)

$V_{0} = \frac{E}{e} - \frac{ϕ}{e}$

$K E_{\max} = E - h ν_{th}$

$K E_{\max} is highest for least ν_{th}$

$As metal (1) having minimum value of work function$

$So X_{1} will have maximum kinetic energy$

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Solution

Q. The graph shows variation of stopping potential V0 with the frequency ν of the incident radiation for three photosensitive metals X1, X2 and X3. Which metal will give out electrons with greater kinetic energy, for the same wavelength of incident radiation [2026]

1 X1

2 X2

3 X3