A metal plate of area 1 × 10 ^ - 4 m ^ 2 is illuminated by a radiation of intensity 16 milliW m ^ 2 . The work function of the metal is 5eV . The energy of the incident photons is 10eV and only 10% of it produces photo

Correct answer: C. 10 ^ 11 and 5eV Einstein's photoelectric equation Maximum KE = (hC)/(λ) - ϕ = 10 - 5 = 5eV Intensity I = ( N _ p )(hC)/(λ) tA ⇒ N _ p t = (IA)/(((hC)/(λ))) and number of electrons per second. = ( N _ p t ) × 10% = ((IA)/((hC)/(λ))) × (1)/(10) = 16 × 10 ^ - 3 × 10 ^ - 4 10 × 1.6 × 10 ^ - 19 × 10 = 10 ^ 11 per sec

JEE Main 201910 JanuaryEvening Shift

Exam: JEE Main
Year: 2019
Date: 10 January
Shift: Evening Shift
Subject: Physics
Chapter: Dual Nature of Matter

JEE Main 2019 · 10 January · Evening Shift · Physics · Dual Nature of Matter

A metal plate of area 1 × 10 ^ - 4 m ^ 2 is illuminated by a...

A metal plate of area

1 \times {10}^{- 4}{m}^{2}

is illuminated by a radiation of intensity

16\dfrac{milliW}{{m}^{2}}

. The work function of the metal is

5eV .

The energy of the incident photons is

10eV

and only

10%

of it produces photo electrons. The number of emitted photo electron per second and their maximum energy, respectively, will be:

\left[1eV = 1.6 \times {10}^{- 19}J\right]

Options

A ${10}^{14}$ and $10eV$
B ${10}^{12}$ and $5eV$
C ${10}^{11}$ and $5eV$
D ${10}^{10}$ and $5eV$

Correct answer

C. ${10}^{11}$ and $5eV$

Step-by-step solution

Einstein's photoelectric equation
Maximum

KE = \dfrac{hC}{\lambda} - ϕ

= 10 - 5

= 5eV

Intensity

I = \dfrac{\left({N}_{p}\right)\dfrac{hC}{\lambda}}{tA} ⇒ \dfrac{{N}_{p}}{t} = \dfrac{IA}{\left(\dfrac{hC}{\lambda}\right)}

and number of electrons per second.

= \left(\dfrac{{N}_{p}}{t}\right) \times 10%

= \left(\dfrac{IA}{\dfrac{hC}{\lambda}}\right) \times \dfrac{1}{10} = \dfrac{16 \times {10}^{- 3} \times {10}^{- 4}}{10 \times 1.6 \times {10}^{- 19} \times 10}

= {10}^{11}

per sec

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