Let the threshold wavelength = λ0 nm = λ0×10–9m then
hc(

1

λ

−

1

λ0

) =

1

2

mv2
Substituting the given three experiments,

hc

10−9

(

1

500

−

1

λ0

) =

1

2

m(2.55×105)2 ... (i)

hc

10−9

(

1

450

−

1

λ0

) =

1

2

m(4.35×105)2 ... (ii)

hc

10−9

(

1

400

−

1

λ0

) =

1

2

m(2.55×105)2 ... (iii)
Dividing equation (ii) by equation (i) we get,

λ0−450

450λ0

×

500λ0

λ0−500

= (

4.35

2.55

)2 or

λ0−450

λ0−500

=

450

500

×(

4.35

2.55

)2
or

λ0−450

λ0−500

=

9

10

×

18.92

6.50

or

λ0−450

λ0−500

= 2.62
or λ0 – 450 = 2.62 λ0 – 1310 or – 450 + 1310 = 2.62 λ0–λ0
or 860 = 1.62 λ0 ⇒ λ0 =

860

1.62

= 530.86 nm ≈ 531 nm
Putting this in equation (iii), we get

h×3×108

10−9

(

1

400

−

1

531

) =

1

2

(9.1×10−31)(5.20×105)2
h × 3 × 1017(

131

531×400

) =

1

2

× 27.04 × 1010 × 9.1 × 10−31
⇒ h = 6.64 × 10−34Js.