Question

Threshold frequency, ${\nu }_0$ is the minimum frequency which a photon must possess to eject an electron from a metal. It is different for different metals. When a photon of frequency $1.0\times {10}^{15}{s}^{-1}$ was allowed to hit a metal surface, an electron having $1.988\times {10}_{-19}J$ of kinetic energy was emitted. Calculate the threshold frequency of this metal. Show that an electron will not be emitted. if a photon with a wavelength equal to $600nm$ hits the metal surface.

Answer 1

Calculate the threshold frequency

Energy of photon = Threshold energy + Kinetic energy

$hv=h{v}_0+KE$

Where, $v_0=$ threshold frequency

$v=$ frequency of the photon striking the surface

According to the question:

Frequency of photon $v=1\times {10}^{15}{s}^{-1}$

Kinetic energy $=1.988\times {10}^{-19}J$

Photoelectric effect: $hv=h{v}_0+KE$

$(6.626\times {10}^{-34}\times 1\times {10}^{15})$

$=(6.626\times {10}^{-34}\times v_o)+(1.988\times {10}^{-19})J$

$v_o=\frac{(6.626\times {10}^{-19})-(1.988\times {10}^{-19})}{6.626\times {10}^{-34}}$

$v_0=7\times {10}^{14}Hz$

Checking whether photon of $600nm$ would eject electron or not.

Wavelength of the given photon $=600nm$

$=600\times {10}^{-9}m$

Frequency of given photon striking metal surface

$v=\frac{c}{\lambda }$

$v=\frac{3\times {10}^0}{600\times {10}^{-9}}$

$=5\times {10}^{14}Hz$

$v=5\times {10}^{14}Hz$

Since $v<v_0(v_0=7\times {10}^{14}Hz)$

Hence, photoelectrons will not be ejected.