Home work semiconductor device
2) The work function for tungsten is 4.58 eV. (a) Find the threshold frequency and wavelength for the
photoelectric effect. Then, find the stopping potential if the wavelength of the incident light is (b)
200 nm & (c) 250 nm.
The answer :
From you know : e = 1.6×10 ⁻ ¹⁹ J
c = 3×108m/s
ϕ = 4.58 eV
h = 6.63 × 10 ⁻ ³⁴ J.s,
ϕ = 4.58 × 1.6 ×10 ⁻¹⁹ J
= 7.32 × 10 ⁻¹⁹ J
a) Threshold wavelength = λt = ( h x c ) : ϕ
= ( 6.63×10 ⁻³⁴ × 3×10⁸ ) : 7.32 × 10 ⁻¹⁹
= 2.717×10 ⁻⁷ m
= 2717 A˚
b) Threshold Frequency = f t = c : λt
= 3×10 ⁸ m/s : 2.71.7×10 ⁻⁹ m
= 1.11 x 10 ⁻¹⁵ Hz
c) Stopping Potential ( S ) when :
#) λ = 200 nm = 200 x 10 ⁻⁹ m
KE max = e x S
KE max = h x f - ϕ
KE max = h x ( c : λ ) – ϕ
KE max = 6.63 × 10 ⁻ ³⁴ J.s x (3 × 10 ⁸ m/s : 200 x 10 ⁻⁹ m ) – 4.58 eV
= 1.63 eV
S = KE max : e
S = 1.63 eV : eV
S = 1.63
#) λ = 250 nm = 250 x 10 ⁻⁹ m
KE max = e x S
KE max = h x f - ϕ
KE max = h x ( c : λ ) – ϕ
KE max = 6.63 × 10 ⁻ ³⁴ J.s x (3 × 10 ⁸ m/s : 250 x 10 ⁻⁹ m ) – 4.58 eV
= 0.39 eV
S = KE max : e
S = 0.39 eV : eV
S = 0.39
Thanks