Question

Calculate the change in energy of an atom that emits a photon of wavelength 2.21 meters. (Planck’s constant is 6.626 x 10-34 joule seconds, the speed of light is 2.998 x 108 m/s)
* 8.9886 x10-26 joules
*4.8844 x 10-42 joules second
* 1.9864 x 10-25 joules
* 1.4643 x 10-33 joules /second

Answer 1

Answer : The correct option is, $8.988\times 10^{-26}joules$

Solution :

Formula used :

$E=\frac{h\times c}{\lambda}$

where,

E = change in energy

h = Planck’s constant = $6.626\times 10^{-34}J/s$

c = speed of light = $2.998\times 10^{8}m/s$

$\lambda$ = wavelength = 2.21 m

Now put all the given values in the above formula, we get the energy of an atom.

$E=\frac{(6.626\times 10^{-34}J/s)\times (2.998\times 10^{8}m/s)}{2.21m}$

$E=8.988\times 10^{-26}joules$

Therefore, the energy of an atom is, $8.988\times 10^{-26}joules$