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Question

The work function of a metal is 1eV. On making light of wavelength $$\displaystyle 3000\mathring {A} $$  incident on this metal, the velocity of photoelectrons emitted from it for photoelectric emission will be 


A
2955km/s
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B
4200km/s
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C
1100km/s
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D
3000km/s
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Solution

The correct option is C $$\displaystyle 1100km/s$$
The given answers are having the dimension of length [L].  It can be solvable.
The photo electric effect can be written as:

$$E=\dfrac{mv^2}{2}+\phi$$

$$\dfrac{mv^2}{2}=E-\phi=\Delta E$$

So, $$v = \sqrt {[2*E/m)]} $$
Now, the work function of a metal is $$1eV=1.6 \times 10^{-19}$$

And energy of the photon is $$\dfrac{hc}{\lambda}=6.6\times 10^{-19}$$

So, $$\Delta E= 5.6\times 10^{-19}$$
Now the velocity of the photo electron is $$v = \sqrt {[2*E/m)]}=1109053 m/s=1100 km/s$$
So, the answer is option (C).

Physics
NCERT
Standard XII

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