The surface of sodium metal is irradiated with radiation of wavelength x nm. The kinetic energy of ejected electrons is 2.8 × 10⁻-20 J. The work function of sodium is 2.3 eV. The value of x is ______ × 102 nm. (Nearest integer)
(Given: h = 6.6 × 10⁻-34 J s; 1 eV = 1.6 × 10-19 J; c = 3.0 × 108 m s⁻¹)
Using Einstein’s photoelectric equation, hν = ϕ + Kmax
Also, hν = hc/λ
So, hc/λ = ϕ + Kmax
Now convert the work function into joule: ϕ = 2.3 eV = 2.3 × 1.6 × 10⁻-19 J = 3.68 × 10⁻-19 J
Given kinetic energy, Kmax = 2.8 × 10-20 J
Hence total photon energy is E = ϕ + Kmax
E = 3.68 × 10-19 + 2.8 × 10-20
E = 3.96 × 10-19 J
Now, λ = hc/E
λ = (6.6 × 10-34 × 3.0 × 108) / (3.96 × 10-19)
λ = 5 × 10-7 m λ = 500 nm x = 5 × 102 nm
Final Answer x = 5
Hence, the required value of x (nearest integer) is 5.
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