Question

plZ ans this qus.. ​

Answer 1

Before solving this question ,first understand these thing

Stopping potential- This is the potential which is required to stop the electron, moving from metallic surface....

Formula of stopping potential={(energy of light,which is incident on metal)}-(threshold energy)}/e or,(hv-hvo)/e=hc(1/wavelength -1/thershold wavelength)/e

Now lets Move on question

I had taken lambda=x and threshold lambda as x0

first case:

3V0=hc/e(1/x-1/x0)

V0=hc/3e(1/x -1/x0).....i)

Second case:

V0=hc/e(1/2x -1/x0).....ii)

equate both the eqaution

hc/3e(1/x -1/x0)=hc/e(1/2x-1/x0)

1/3x-1/3x0=1/2x-1/x0

1/3x-1/2x=-1/x0+1/3x0

-1/6x=-3+1/3x0

-1/6x=-2/3xo

x0=4x

now,put x0=threshold wavelength and x=wavelength of light

hence

the threshold Lambda equals to 4 x wavelength of light, which is date on it in first case,

{hope it helps}