The electric field of light wave is given as \overrightarrow E = 10^ - 3\cos ( 2\pi x \over 5 \times 10^ - 7 - 2\pi \times 6 \times 10^14t )\mathop x\limits^ \wedge \rm N \over C This light falls on a metal plate of work function 2eV. The stopping potential of the photoelectrons is : Given, E (in eV) = 12375/ \lambda (inÅ)

Dual Nature of Radiation PYQ — JEE Main 2019 · 9 Apr · Shift 1 · Q60 with Solution

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Correct answer: B

The electric field is

\vec E = 10^{-3}\cos\left(\frac{2\pi x}{5\times 10^{-7}} - 2\pi\times 6\times 10^{14} t\right)\hat x

Compare with the standard form:

E = E_0 \cos\left(\frac{2\pi x}{\lambda} - 2\pi \nu t\right)

So,

\lambda = 5\times 10^{-7}\ \text{m}

and

\nu = 6\times 10^{14}\ \text{Hz}

Since

1\ \text{m} = 10^{10}\ \text{\AA}

we get

\lambda = 5\times 10^{-7}\times 10^{10} = 5\times 10^3\ \text{\AA} = 5000\ \text{\AA}

Given,

E(\text{in eV}) = \frac{12375}{\lambda(\text{in \AA})}

Therefore,

E = \frac{12375}{5000} = 2.475\ \text{eV}

Work function of metal:

\phi = 2\ \text{eV}

Maximum kinetic energy of emitted photoelectrons:

K_{\max} = E - \phi = 2.475 - 2 = 0.475\ \text{eV}

Since

K_{\max}(\text{in eV}) = eV_s

when expressed numerically in eV,

V_s = 0.475\ \text{V} \approx 0.48\ \text{V}

Hence, the correct option is B.