KEAM 2015 Physics and Chemistry Paper

According to Einstein's photoelectric equation, i.e.
E=W0+Kmax
where, ‌‌Kmax=‌

1

2

mvmax2= maximum kinetic energy of emitted electrons.
Given,
λ1=400nm=400×10−9m
λ2=250nm=250×10−9m
In the first condition,
‌

hc

400×10−9

=‌

1

2

mv2+ϕ...(i)
In the second condition,
‌

hc

250×10−19

=‌

1

2

m(2v)2+ϕ
‌

hc

250×10−19

=‌

1

2

m4v2+ϕ...(ii)
From the Eq. (i), we get
‌

1

2

mv2=‌

hc

400×10−9

−ϕ....(ii)
Putting the value of Eq. (iii) in Eq. (ii), we get
⇒3ϕ=‌

4hc

400×10−9

−‌

hc

250×10−9

⇒3ϕ=‌

hc

100×10−9

−‌

hc

250×10−9

⇒3ϕ=hc×109(‌

1

100

−‌

1

250

)
⇒3ϕ=hc×109(0.01−0.004)
∴‌‌3ϕ=6hc×106
Work function of the metal,
ϕ=2hc×106J