Question

If the electron in H-atom radiates a photon of wavelength
4860 A, the KE of the electron:
(a) decreases by 2.0x10-12
(b) increases by 4.1% 10-240
(c) decreases by 4.1 x 10-19
(d) increases by 8.2 x 10-1° J​

Answer 1

Explanation:

if the Wavelength is 4860A°

4860×10-10m

masa of electron is 9.1× 10^-31kg

h = 6.63×10^-34 js

the KE will be

KE is = 1/2 mv^2

1/2mv^2 = p^2/2m

so, p= √2mk

and Wavelength = h/p

so, put p value in Wavelength

then Wavelength = h/√2mk

here 'k' is kinetic energy

then put all values in it

then k= h^2/2m (wavelength) ^2

k= (6.63×10^-34) ^2 /2×9.1×10^-31 (4860× 10^-10) ^2

43.96×10^68/18.2×10^-10 (23619600×10^20)

43.96×10^68/429×10^6

0.10247×10^62