Test Index

BITSAT 2012 Paper

Section: Physics

Question : 36 of 150

Marks: +1, -0

Solution:

Energy of incident radiation is given by:

E = \frac{hc}{\lambda}

= \frac{6.64 \times 10^{-34} \times 3 \times 10^{8}}{300 \times 10^{-9}}

= 6.64 \times 10^{-19} = 4.05 \text{ eV}

The work function of the metal is equal to

2.54 \text{ eV}

Hence the maximum kinetic energy of an ejected photoelectron would be

1.59 \text{ eV}

And the stopping potential needed to stop this would be

1.59 \text{ V}

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