KEAM 2021 Physics and Chemistry Solved Paper

Given, φA=3eV,φB=4eV
(KE)A=(KE)B=1eV,λA=500nm
According to Einstein's photoelectric equation, maximum kinetic energy of photoelectrons,
KE=‌

hc

λ

−φ
For material A,
(KE)A=‌

hc

λA

−φA
Similarly, for material B,
(KE)B=‌

hc

λB

−φB
As, (KE)A=(KE)B
⇒‌‌‌

hc

λA

−φA=‌

hc

λB

−φB
⇒‌‌‌

hc

λA

−‌

hc

λB

=φA−φB
hc(‌

1

λA

−‌

1

λB

)=(3−4)eV
⇒‌

1

λA

−‌

1

λB

=‌

−1×1.6×10−19

6.626×10−34×3×108

‌

1

λB

=‌

1

500×10−9

+‌

1.6×10−19

6.626×10−34×3×108

=2×106+8.05×105
=2×106+8.05×105
=2.805×106
or λB‌‌=3.56×10−7
‌‌=350nm