A difference of 2.3 eV separates two energy levels in an atom. Whatis the frequency of radiation emitted when the atom make atransition from the upper level to the lower level?
Question
A difference of 2.3 eV separates two energy levels in an atom. Whatis the frequency of radiation emitted when the atom make atransition from the upper level to the lower level?
Solution
To find the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level, we can use the equation:
E = hf
where E is the energy difference between the two levels (2.3 eV), h is Planck's constant (6.626 x 10^-34 J*s), and f is the frequency of the radiation.
First, we need to convert the energy difference from electron volts (eV) to joules (J). Since 1 eV is equal to 1.602 x 10^-19 J, we can calculate:
E = 2.3 eV * (1.602 x 10^-19 J/eV) E = 3.686 x 10^-19 J
Now, we can rearrange the equation to solve for the frequency:
f = E / h f = (3.686 x 10^-19 J) / (6.626 x 10^-34 J*s) f = 5.57 x 10^14 Hz
Therefore, the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level is approximately 5.57 x 10^14 Hz.
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