JEE Main 2019 — Physics solution

JEE Main2019Physics

The electric field of light wave is given as $E = 1 0^{- 3} cos (\frac{2 π x}{5 \times 10 ^{- 7}} - 2 π \times 6 \times 10^{14} t) \overset{x}{^} \frac{N}{C}$ . This light falls on a metal plate of work function 2eV. The stopping potential of the photoelectrons is : Given, E (in eV) = $\frac{12375}{λ ( in A 0 )}$

A
0.48 V
B
2.48 V
C
0.72 V
D
2.0 V

Solution

$λ = 5 \times 1 0^{- 7} m v = \frac{3 \times 10 ^{8}}{5 \times 10 ^{- 7}} = 6 \times 1 0^{14} H z E = \frac{12375}{5000} = 2.475 e V K_{m a x} = E - ϕ = 0.48 e V$

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