Question

When an electron of an atom falls from a higher energy level to the ground state, the atom loses 9.4145 x 10-25 joules of energy. What is the wavelength of the radiation emitted as a result of this transition? (Planck’s constant is 6.626 x 10-34 joule seconds; the speed of light is 2.998 x 108m/s)?

Answer 1

Answer:

The wavelength of the radiation emitted as a result of this transition is 211 mm.

Explanation:

Energy loose by the atom = $E=9.4145\times 10^{-25} Joules$

Relationship between wavelength and energy is given by photoelectric equation:

$E=\frac{hc}{\lambda }$

$\lambda$ = wavelength of the radiation

h = Planck's constant

c = speed of light

$\lambda =\frac{hc}{E}=\frac{6.626 x 10-34 Joule seconds\times 2.998\times 10^8 m/s}{9.4145\times 10^{-25} Joules}=0.211=211 mm$

The wavelength of the radiation emitted as a result of this transition is 211 mm.

Answer 2

$E = \frac{hc}{wavelength}$
given E = 9.4145E-25
h = 6.626E-34
c = 2.998E8
sub values into the equation above, and solve for wavelength.
You will get 0.211m