Question

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

Answer 1

Assuming the threshold wavelength to be `lambda_0 " nm"(= lambda_0 xx 10^(-9)" m")`, the kinetic energy of the radiation is given as:

`"h" ("v" - "v"_0) = 1/2 " mv"^2` ...........(i)

Since v = `"c" / λ`, the equation (i) can write the following types-

`"hc" [1/λ - 1/λ_0] = 1/2 "mv"^2` ..............(ii)

Upon replacing the results given by the three experiments in the equation (ii),

`"hc"/10^-9 [1/500 - 1/λ_0] = 1/2 "m" xx (2.55 xx 10^6)^2` .......(iii)

`"hc"/10^-9 [1/450 - 1/λ_0] = 1/2 "m" xx (4.35 xx 10^6)^2` .....…(iv)

`"hc"/10^-9 [1/400 - 1/λ_0] = 1/2 "m" xx (5.35 xx 10^6)^2` ........…(v)

Dividing equation (iv) by equation (iii)

`(λ_0 - 450)/(450 λ_0) xx (500 λ_0)/(λ_0 - 500) = [4.35/2.55]^2`

or `(λ_0 - 450)/(λ_0 - 500) = [4.35/2.55]^2 xx 450/500` = 2.619

or λ₀ − 450 = 2.619 λ₀ − (500 × 2.619)

or 1.619 λ₀ = 1309.5 − 450 = 859.5

or λ₀ = `859.5/1.619` = 530.88 = 531 nm

The value of the Planck’s constant is H, the value of the λ₀ can be obtained on replacing in one of the three classes.

Answer 2

Let the threshold wavelength to be λ₀ nm = λ₀×10^-9 m . 

Following equation holds true for photoelectric emission in given case:

K.E. = 1/2 mv² = h(ν - ν₀)

⇒1/2 mv² = hν - hν₀

⇒ hν₀ = hν - 1/2 mv²

⇒ hc/λ₀ = hc/λ - 1/2 mv²

`=> lambda_0 = 1/(1/lambda - 1/2 m/hc v^2) = 1/(1/lambda - 1/2 (9.1xx10^(-31))/(6.6xx10^(-34)xx3xx10^8))v^2`

a) Substituting the value of λ and v from the above given data, we get three values of λ₀ as,
λ₀₍₁₎ = 541 nm
λ₀₍₂₎ = 546 nm
λ₀₍₃₎ = 542 nm

Threshold frequency = λ_av = {λ₀₍₁₎+λ₀₍₂₎+λ₀₍₃₎}/3 = (541+546+542)/3 = 543 (approx 540)