Answer:
E = 3.77×10⁻¹⁹ J
Explanation:
Given data:
Wavelength of absorption line = 527 nm (527×10⁻⁹m)
Energy of absorption line = ?
Solution:
Formula:
E = hc/λ
h = planck's constant = 6.63×10⁻³⁴ Js
c = speed of wave = 3×10⁸ m/s
by putting values,
E = 6.63×10⁻³⁴ Js × 3×10⁸ m/s / 527×10⁻⁹m
E = 19.89×10⁻²⁶ Jm /527×10⁻⁹m
E = 0.0377×10⁻¹⁷ J
E = 3.77×10⁻¹⁹ J
Answer:
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Answer : The energy of the photon emitted is, -12.1 eV
Explanation :
First we have to calculate the
orbit of hydrogen atom.
Formula used :

where,
= energy of
orbit
n = number of orbit
Z = atomic number of hydrogen atom = 1
Energy of n = 1 in an hydrogen atom:

Energy of n = 2 in an hydrogen atom:

Energy change transition from n = 1 to n = 3 occurs.
Let energy change be E.

The negative sign indicates that energy of the photon emitted.
Thus, the energy of the photon emitted is, -12.1 eV