MHTCET 13 May 2023 Shift 2 Solved Paper

From Einstein's photoelectric equation,
‌KEmax‌=hv−hv0
or eV‌=hv−hv0
⇒‌‌eV‌=‌

hc

λ

−‌

hc

λ0

Here, V is stopping potential and λ is the incident wavelength, whereas λ0 its threshold wavelength.
For case 1e(4.8)=‌

hc

λ

−‌

hc

λ0

.... (i)
For case 2e(1.6)=‌

hc

2λ

−‌

hc

λ0

..........(ii)
From Eqs. (i) and (ii), we get
‌e×(3.2)=‌

hc

2λ

⇒‌‌e(1.6)=‌

hc

4λ

‌‌....‌‌.... (iii)
Putting the Eq. (iii) in Eq. (ii), we get
‌⇒‌

hc

4λ

=‌

hc

2λ

−‌

hc

λ0

‌⇒‌

hc

λ0

=‌

hc

2λ

−‌

hc

4λ

‌⇒‌‌‌

hc

λ0

=‌

hc

4λ

‌⇒‌‌λ0=4λ