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The work functions of Silver and Sodium are 4.6 and 2.3 \mathrm{eV}, respectively. The ratio of the slope of the stopping potential versus frequency plot for Silver to that of Sodium is:
 

Option: 1

1:1


Option: 2

1:2


Option: 3

2:1


Option: 4

1:4


Answers (1)

best_answer


According to Einstein's photoelectric equation \mathrm{K}_{\max }=h v-\phi_0

Where the symbols have their usual meaning,

But \mathrm{K}_{\max }=\mathrm{eV}_{\mathrm{s}} where \mathrm{V}_{\mathrm{s}} is the stopping potential \mathrm{V}_{\mathrm{s}}=\frac{\mathrm{h}}{\mathrm{e}} \mathrm{v}-\frac{\phi_0}{\mathrm{e}}

Thus, the graph between \mathrm{V}_{\mathrm{s}}-\mathrm{v} is a straight line. Compare the above relation with \mathrm{y}=\mathrm{mx}+\mathrm{C} 

\therefore Slope of \mathrm{V}_{\mathrm{s}}-\mathrm{v} graph = \frac{\mathrm{h}}{\mathrm{e}}

It is same for both the metals.
\therefore  Ratio of the slopes =1.

Posted by

Sanket Gandhi

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