A 0.1 kg magnet is floating above a second magnet. The magnetic force on the 0.1 kg magnet is
2 answers:
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Answer:
(a) The energy of the photon is 1.632 x J.
(b) The wavelength of the photon is 1.2 x m.
(c) The frequency of the photon is 2.47 x Hz.
Explanation:
Let;
= -13.60 ev
= -3.40 ev
(a) Energy of the emitted photon can be determined as;
-
= -3.40 - (-13.60)
= -3.40 + 13.60
= 10.20 eV
= 10.20(1.6 x )
-
= 1.632 x
Joules
The energy of the emitted photon is 10.20 eV (or 1.632 x Joules).
(b) The wavelength, λ, can be determined as;
E = (hc)/ λ
where: E is the energy of the photon, h is the Planck's constant (6.6 x Js), c is the speed of light (3 x
m/s) and λ is the wavelength.
10.20(1.6 x ) = (6.6 x
* 3 x
)/ λ
λ =
= 1.213 x
Wavelength of the photon is 1.2 x m.
(c) The frequency can be determined by;
E = hf
where f is the frequency of the photon.
1.632 x = 6.6 x
x f
f =
= 2.47 x Hz
Frequency of the emitted photon is 2.47 x Hz.