COMEDK Exam 2018 Solved Paper

According to first situation wavelength of incident radiation,
λ1=λ
Stopping potential, V01=3VS
According to Einstein's photoelectric equation,
⇒‌‌Kmax‌=hv1−φ
‌‌eV01‌=‌

hc

λ1

−‌

hc

λ0

[where, λ0= threshold wavelength.]
⇒e⋅3Vs=hc(‌

1

λ

−‌

1

λ0

)
⇒3Vse=hc(‌

1

λ

−‌

1

λ0

) . . . (i)
According to second situation, wavelength of incident radiation,
λ2=2λ
Stopping potential, V02=VS
Again, Einstein's photelectric equation,
Kmax‌=hv2−φ
⇒‌‌eV02‌=hc(‌

1

λ2

−‌

1

λ0

)
⇒‌‌eVs‌=hc(‌

1

2λ

−‌

1

λ0

) . . . (ii)

Dividing Eq. (i) by Eq (ii), we get
‌3=‌

‌ 1 λ − 1 λ0

1 2λ −‌ 1 λ0

⇒‌‌‌

3

2λ

−‌

3

λ0

=‌

1

λ

−‌

1

λ0

⇒‌‌λ0‌=4λ