The electric field associated with a light wave is $E = E_{0} sin (1.57 \times 10^{7} (x - c t))$ , where $x$ is in meter and $t$ is in second. If this light is used to produce photoelectric emission, from the surface of a metal has a work function $1.9 e V$ , then the stopping potential will be ?

1.2 V

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1.5 V

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1.75 V

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1.9 V

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Solution

The correct option is A $1.2 V$
Comparing the the given expression for electric filed with the standard expression, $E = E_{0} s i n (k x - ω t)$
we get $k = 1.57 \times 10^{7} m^{- 1}$ where $k$ is propagation constant which is related to wavelength, $λ$ as
follows $k = \frac{2 π}{λ}$ so putting value of $k$ we get $λ = 4 \times 10^{- 7} m = 4000 A n g s t r o m$
So energy of incident photon will be $E = \frac{h c}{λ} = \frac{12400}{4000} e v = 3.1 e V$
Out of which $1.9 e V$ will be consumed against $w o r k$ $f u n c t i o n, W = 1.9 e V$ so maximum kinetic energy $K E_{m a x} = 3.1 e V - 1.9 e V = 1.2 e V$
So the stopping potential will be $V_{0} = \frac{K E_{m a x}}{c h a r g e} = \frac{1.2 e V}{e} = 1.2 V$

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The electric field associated with a light wave is E=E0sin(1.57 ×107(x−ct)), where x is in meter and t is in second. If this light is used to produce photoelectric emission, from the surface of a metal has a work function 1.9 eV, then the stopping potential will be ?

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