MHT CET 2021 Physics Solved Paper

(c) According to Einstein's photoelectric equation, maximum energy of emitted electrons is given by
Kmax‌‌=‌

1

2

mvmax2=hv−hv0
⇒‌‌‌

1

2

mvmax2‌‌=h(v−v0)‌‌‌⋅⋅⋅⋅⋅⋅⋅(i)
When, frequency of incident radiation, becomes 2v, then velocity of emitted electron is v1, hence from Eq. (i), we get
‌

1

2

mv12=h(2v−v0)
‌

1

2

mv12=hv0‌‌‌⋅⋅⋅⋅⋅⋅⋅(ii)
Again, when frequency becomes 5v0, the velocity of emitted electron is v2.
∴ From Eq. (i), we have
‌

1

2

mv22=h(5v−v)
⇒‌‌‌

1

2

mv22=4hv
Dividing Eq. (ii) by Eq. (iii), we get
‌‌

‌ 1 2 mv12

1 2 mv22

=‌

hv

4hv

⇒‌‌(‌

v1

v2

)2=(‌

1

2

)2
⇒‌‌v1:v2=1:2