AP EAPCET 08-Jul-2022 Shift 1 Paper

Maximum speed v_{max} of emitted electrons is given by

1

2

mvmax2=

hc

λ

−ϕ0
In given experiment we have,

1

2

mv12=

hc

(600×10−9)

ϕ0 ........(i)
and

1

2

mv22=

hc

(200×10−9)

−ϕ0 ........(ii)
Given, v1=

1

3

v2,As‌V2>v1
Substituting for v_2 in eq. (ii) we get,

1

2

m(3v1)2=

hc

200×10−9

−ϕ0
or 9(

1

2

mv12)=

hc

200×10−9

−ϕ0
Substituting for

1

2

mv12 from eq.(i) in above
we get,
9(

hc

600×10−9

)−ϕ0=

hc

200×10−9

−ϕ0
⇒

hc

10−7

(

9

6

−

1

2

)=8ϕ0
Work‌function,ϕ0=

hc

8

×107J