Question

Question 2.52 Following results are observed when sodium metal is irradiated with different wavelengths. Calculate

(a) threshold wavelength and,

(b) Planck’s constant.




λ (nm) 500 450 400




v × 10–5 (cm s–1) 2.55 4.35 5.35

Class XI Structure of Atom Page 68

Answer 1

Answer

(a) Let us suppose threshold wavelength to be λDnm (= λ0 x 10-9 m), the kinetic energy of the radiation is given as:

h(v-v0)=1/2 mv2

Or

hc(1/λ - 1/λD) = 1/2 mv2

.......see the file.......

Threshold wavelength (λD)= 540 nm

 Substituting this value in equation iii,we get

= h  x (3x108)/10-9 [1/400-1/540]

= [(9.11x10-31)(5.20x106)2] / 2

=6.66x10-34Js

Answer 2

We know,
Kinetic energy { K.E } = h( v - v₀ ) = 1/2 mv²
and ν = c/λ
so,kinetic energy { K.E } = hc(1/λ - 1/λ₀ ) = 1/2mv²

now, putting λ and v values from given data .

hc{1/500 - 1/λ₀ ) = 1/2 m(2.55)²___(1)
hc{1/450 - 1/λ₀ ) = 1/2 m(4.35)²___(2)
hc{1/400 -1/λ₀ ) = 1/2 m(5.35)² ___(3)

now, from equations (1) and (2)
dividing equation (2) by (1)
{1/500 - 1/λ₀ }/{1/450 - 1/λ₀ } = (2.55/4.35)²
{1/500 - 1/λ₀ }/{1/450 - 1/λ₀ } = 0.343
{1/500 - 1/λ₀ } = 0.343/450 - 0.343/λ₀
λ₀ ≈ 531 nm

now, from equation (3)
h× 3 × 10^8{1/500×10^-9 - 1/531 × 10^-9}=1/2 × 9.11 × 10^-31 × (5.35)²
h = 6.6539 × 10^-34 Js