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Light of wavelength 0.6 mum from a sodiu...

Light of wavelength 0.6 `mum` from a sodium lamp falls on a photocell and causes, the emission of photoelectrons for which the stopping potential is 0.5 V. With light of wavelength 0.4 `mum` from a sodium lamp, stopping potential is 1.5V. With this data, the value of h/e is `(nxx10^(-15))` Vs. Find value of n.

A

`0.75 eV`

B

`1.5eV`

C

`3eV`

D

`2.5eV`

Text Solution

Verified by Experts

The correct Answer is:
B
`E-W_0=(1)/(2)mv^2=eV_s`
or `(hc)/(lamda)-W_0=eV_s`
Hence, `(hc)/(0.4xx10^-6)-W_0=e(1.5)`
Solving, we get `W_0=1.5eV`
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