Question

The energy difference between two electronic states of hydrogen atom is 245.9 kJ
mol-1. Calculate the wavelength of light emitted when an electron drops from the
higher to the lower state. (Planck's constant, h =3.99 10-13kJ sec. mol-l).

answer is 4.87 into 10-7 m​

Answer 1

Answer:

ANSWER

ΔE=E

2

−E

1

=hv

∴214.68=39.79×10

−14

×v

v=5.39×10

14

cps

Answer 2

ANSWER:

ANSWER

ΔE= E2−E1=hv

∴214.68=39.79×10

−14×v

v=5.39×10 14 cps