Representing the stopping potential V along y-axis and `((1)/(lamda))` along x-axis for a given photocathode, the curve is a straight line, the slope of which is equal to

Options

A. `(e)/(hc)`
B. `(hc)/(e)`
C. `(ec)/(h)`
D. `(e)/(W)`

Answer

B

Explanation

The maximum KE of the photoelectron is given by
`((1)/(2)mv^2)_(max)=hv-W`
Now, `v=(c)/(lamda)` and `((1)/(2)mv^2)=eV`
`eV=(hc)/(lamda)-W`
or `V=((hc)/(e))(1)/(lamda)-(W)/(e)`
Slope of straight line `=(hc)/(e)`
Intercept of straight line `=-((W)/(e))`