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light quanta with a energy 4.9 \mathrm{eV} eject photo electrons from metal with work function 4.5 \mathrm{eV}. Find the maximum impuse transmitted to the surface of the metal when each electrons flies out.

Option: 1

2.45 \times 10^{-25} \mathrm{~kg} \mathrm{~m} / \mathrm{s}


Option: 2

2.95 \times 10^{-25} \mathrm{~kg}-\mathrm{m} / \mathrm{s}


Option: 3

3.15 \times 10^{-25} \mathrm{~kg} \mathrm{~m} / \mathrm{s}


Option: 4

3.45 \times 10^{-25} \mathrm{~kg} \mathrm{~m} / \mathrm{s}


Answers (1)

best_answer

According to  photo electric equation
E=\frac{1}{2} m V_{\max }^2=h \nu-\phi=4.9-4.5=0.4 \mathrm{eV}
If E  be the energy of each ejected photo electron, momentum of electron is


p=\sqrt{2 m E}

we know that a change of momentum is impulse. Hence the whole momentum of electron is gained when it is ejected out thus impulse on surface is

Impulse=\sqrt{2 m E}

Maximum Impulse=\sqrt{2 \times 9.1 \times 10^{-31} \times 0.4 \times 1.6 \times 10^{-19}} \\
                             =3.45 \times 10^{-25} \mathrm{~kg}-\mathrm{m} / \mathrm{s} .

Posted by

Pankaj Sanodiya

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