From Einstein photoelectric equation, E=ϕ0+KEmax For metal A4=ϕA+TA .........(i) For metal B4.5=ϕB+(TA−1.5) .........(ii) From Eqs. (i) and (ii), we get ϕB−ϕA=2 Now, according to de-Broglie hypothesis, λA=
h
mv
=
h
√2mTA
Similarly, λB=
h
√2mTB
∴
λA
λB
=√
TB
TA
=√
TA−1.5
TA
=(1−
1.5
TA
)1∕2 (
1
2
)2=1−
1.5
TA
On solving, TA=2.0eV So, ϕA=4−TA=4−2=2.0eV ϕB=6−TA=6−2=4.0eV