JEE Main 25 Feb 2021 Shift 2 Solved Paper

Given, stopping potential (V1)=0.710V
Incident wavelength of electrons (λ1)=491nm
=491×10−9m
Let λ2 will be the new incident wavelength of electrons.
Stopping potential (V2)=1.43V
As, energy (E)=‌

1240

λ1

=φ0+eV
where, φ0 is work-function and V is applied potential
∴‌‌E1=‌

1240

491

=φ0+0.71 - 1.43 . . . (i)
and E2=‌

1240

λ2

=φ0+1.43. . . (ii)
Now, subtracting Eqs. (i) from (ii), we get
‌E2−E1‌‌=1240(‌

1

λ2

−‌

1

491

)=0.72
⇒‌‌

1

λ2

‌‌=0.00058+0.00204=0.00262
∴‌λ2‌‌=381.7nm
‌λ2‌‌=382nm