Light quanta with energy 4.9 eV eject photoelectrons from metal with work function 4.5 eV. Find the maximum impulse transmitted to the surface of the metal when each electron flies out.

Explanation

According to Einstein's photoelectric equation,
`(1)/(2)mv^(2)_(max)=hc-W`
`=4.9-4.5=0.5eV`
If E be the enegy , then
`E=(1)/(2)mv^(2)` or `v=sqrt((2E)/(m))`
Momentum `=mv=sqrt(2Em)`
We know that change of momentum is impulse Hence, impulse `=mv-0`
Impulse`=mv=sqrt(2Em)`
Substituting the values, we get Maximum impulse
`=sqrt({2xx(0.4xx1.6xx10^(-19))xx9.1xx10^(-31)})`
`=3.45xx10^(-25)ms^(-1)`