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Dual Nature of Radiation question

2023 · 6 Apr · Shift 2 · Q55

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Dual Nature of Radiation
2023 · 6 Apr · Shift 2 · Q55

Dual Nature of Radiation question

2023 · 6 Apr · Shift 2 · Q55

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1

The work functions of Aluminium and Gold are

4.1 ~\mathrm{eV}

and and

5.1 ~\mathrm{eV}

respectively. The ratio of the slope of the stopping potential versus frequency plot for Gold to that of Aluminium is

A
1.5
B
1.24
C
1
D
2

View written solutionFree

Correct answer: C

For the photoelectric effect, Einstein’s equation is

$eV_s = h\nu - \phi$

where:

$V_s$ = stopping potential
$\nu$ = frequency of incident light
$\phi$ = work function
$h$ = Planck’s constant
$e$ = electronic charge

Rearranging in the form of $V_s$ versus $\nu$ :

$V_s = \frac{h}{e}\nu - \frac{\phi}{e}$

This is a straight-line equation of the form:

$y = mx + c$

So, the slope of the graph of stopping potential versus frequency is

$m = \frac{h}{e}$

Notice that the slope depends only on $h$ and $e$ , which are universal constants. It does not depend on the work function $\phi$ of the metal.

Therefore, both Aluminium and Gold have the same slope.

Hence, the required ratio is

$\frac{\text{slope for Gold}}{\text{slope for Aluminium}} = \frac{h/e}{h/e} = 1$

Therefore, the correct option is:

$\boxed{\text{C: }1}$

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