Question

A metal plate of area 1 × 10^-4 m² is illuminated by a radiation of intensity 16 m W/m². The work function of the metal is 5 eV. The energy of the incident photons is 10 eV and only 10% of it produces photoelectrons. The number of emitted photoelectrons per second and their maximum energy, respectively will be (Take, 1 eV = 1.6 × 10^-19 J)
1. 10¹¹ and 5 eV
2. 10¹² and 5 eV
3. 10¹⁰ and 5 eV
4. 10¹⁴ and 10 eV

Answer 1

Correct Answer - Option 1 : 10¹¹ and 5 eV

Concept:

We know that, intensity of a radiation I with energy ‘E’ incident on a plate per second per unit area is given as,

\(I = \frac{{dE}}{{dA \times dt}}\)

\(\Rightarrow \frac{{dE}}{{dt}} = IdA\;or\;IA\)

i.e., energy incident per unit time = IA

Using Einstein’s photoelectric equation, we can find kinetic energy of the incident radiation as

\(E = \frac{1}{2}m{v^2} + \phi \) 

Energy per unit time for incident photons will be

∵ E = Nhv

Calculation:

Given,

Area of the metal plate, A = 1 × 10^-4 m²

Intensity of the radiation, I = 16 m W/m²

Work function of the metal, ϕ = 5 eV

Energy of the incident photons E = 10 eV

1 eV = 1.6 × 10^-19 J

Substituting the given values, we get

\(\frac{{dE}}{{dt}} = 16 \times {10^{ - 3}} \times 1 \times {10^{ - 4}}\) 

\(\frac{{dE}}{{dt}} = 16 \times {10^{ - 7}}W\)      ----(1)

Using Einstein’s photoelectric equation, we can find kinetic energy of the incident radiation as

\(E = \frac{1}{2}m{v^2} + \phi \) 

(Here, ϕ is work function of metal)

E = KE + ϕ

KE = E – ϕ

KE = 10 eV – 5 eV

KE = 5 eV       ----(2)

Now, energy per unit time for incident photons will be

∵ E = Nhv

\(\therefore \frac{{dE}}{{dt}} = hv\frac{{dN}}{{dt}}\;or\;hv\;N'\)       ----(3)

From equations (1) and (3), we get

hvN' = 16 × 10^-7

EN' = 16 × 10^-7

But E = 10 eV and 1 eV = 1.6 × 10^-19 J

So, N' (10 × 1.6 × 10^-19) = 16 × 10^-7

\(\;{N'} = \frac{{16 \times {{10}^{ - 7}}}}{{10 \times 1.6 \times {{10}^{ - 19}}}}\) 

\(N' = \frac{{{{10}^{ - 6}}}}{{{{10}^{ - 18}}}}\) 

⇒ N' = 10¹²

∴ Only 10% of incident photons emit electrons. So, emitted electrons per second are

\(\frac{{10}}{{100}} \times {10^{12}} = {10^{11}}\) 

Therefore, the number of emitted photoelectrons per second and their maximum energy, respectively will be 10¹¹ and 5 eV.