Question

When electromagnetic radiation of wavelength 300nm falls on the surface of sodium,electrons are emitted with a kinetic energy of 1.68 x 105Jmol-l. What is the minimum energy needed to remove an electron from sodium? What is the maximum wavelength that will cause a photoelectron to be emitted?​

Answer 1

$\huge {\orange {\bf {\underline {\underline {Sólutíon}}}:-}}$

The energy (E) of a 300nm photon is given by

$\sf E = hv = \frac{hc}{ \lambda} = \frac{6.626 \times {10}^{ - 34} Js \times 3.0 \times {10}^{8}ms {}^{ - 1} }{300 \times {10}^{ - 9}m }$

The energy of one mole of photons

$\sf = 6.626 \times {10}^{ - 19} J \times 6.022 \times {10}^{23} {mol}^{ - 1}$

$\sf = 3.99 \times {10}^{5} J {mol}^{ - 1}$

The minimum energy needed to remove one mole of electrons from sodium.

$\sf = (3.99 - 1.68) {10}^{5} Jmol {}^{ - 1}$

$\sf = 2.31 \times {10}^{5} Jmol {}^{ - 1}$

The minimum energy for one electron

$\sf = \frac{2.31 \times {10}^{5} Jmol {}^{ - 1} }{6.022 \times {10}^{23} electrons \: mol {}^{ - 1} }$

$\sf = 3.84 \times {10}^{ - 19} J$

This corresponds to the wavelength

$\sf \therefore \lambda = \frac{hc}{E} = \frac{6.626 \times {10}^{ - 34} js \times 3.0 \times {1}^{5} {ms}^{ - 1} }{3.84 {10}^{ - 19}J }$

517nm this wavelength corresponds to green colour in visible spectrum.