Question

when electromagnetic radiation of 300 mm falls on the surface of sodium ,electrons are emitted with a kinetic energy of 1.68*10^5 J/mol.What is the minimum energy needed to remove an electron from Na.what is it's threshold energy​

Answer 1

Answer:

517m

Explanation:

Wavelength of photon (λ)=300nm=300×10−9m

Energy of photon

(hv) = hcλ =(6.626×10−34Js)×(3×108ms−1)(300×10−9m)

= 6.626×10−19J

Energy of one mole of photons

(E) = (6.626×10−19J)×(6.022×1023mol−1)

= 3.99×105Jmol−1

Kinetic energy (K.E.)=1.68×105Jmol−1

The minimum energy needed or threshold energy hv∘ may be calculated as follows :

hv∘ = hv−K.E.

= (3.99×105−1.68×105)Jmol−1

= 2.31×105Jmol−1

Minimum energy needed to remove one electron

= (2.31×105Jmol−1)(6.022×1023mol−1)

= 3.84×10−19J

Wavelength corresponding to this energy

(λ) = hcE

= (6.626×10−34Js)×(3×108ms−1)(3.84×10−19J)

= 5.17×10−7m

= 517m