Question

When electromagnetic radiation of wavelength 300nm falls on the surface of
sodium,electrons are emitted with a kinetic energy of 1.68 x 10⁵ J/mol. 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

AnSwer :-

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

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

The energy of one mole of photons

= 6.626 × 10$^{-19}$J × 6.022 × 10²³ mol$^{-1}$

= 3.99 × 10⁵ J/mol

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

= (3.99 - 1.68)10⁵ J/mol

= 2.31 × 10⁵ J/mol

∴  minimum energy required to remove 1electron from Na atom

The minimum energy for one electron

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

= 3.84 × 10$^{-19}$J

Now,

The corresponds to the wavelength

$\sf \lambda = \dfrac{hc}{E} = \dfrac{6.626 \times {10}^{ - 34}Js \times 3.0 \times {10}^{8} {ms}^{ - 1} }{3.84 \times {10}^{ - 19}J }$

= 517nm

This wavelength corresponds to green colour in visible spectrum .