Answers from problem # 19

(a)
Given;
Spectral lines for hydrogen in the ultra violet region of the electromagnetic spectrum by lyman.
Required;
n₁.
Solution.
n₁≡ the first line in lyman series which in this case is 2.
.: In this series, the value of n₁=2.

(b)
Given;
The lyman series.
Required;
The energy of a line in the lyman series E₁, for n₁=1 and E₂, for n₂=∞
Solution.
From the energy expression;
E = - 13.6eV
            n²
Where; 1eV=1.6E-19J
Case I
n=n₁=1.
⇒E₁= - 13.6×1.6E-19J
                   1²
E₁= - 2.18E−18J
Case II
n=n₂=∞ (at infinity)
⇒E₂= - 13.6×1.6E-19J
                    ∞²
E₂= - 0(zero)J.
.: The energy of a line in the lyman series E₁= -2.18E−18J for n₁=1 and E₂= 0(zero)J for n₂=∞


(c)
Given;
Light emitted of wavelength, λ=1.315μm=(1.315E-6m).
Required;
The frequency, f of this light and the energy per photon, E.
Solution.
From the Planck's energy expression;
E=hc
      λ.
But: c =fλ.
So f =c/λ
Where c is constant speed of light=3.0E8m/s
⇒f = 3.0E8m/s
        1.315E-6m
 f=2.281E14 s-¹.

From energy expression above;
⇒E=hf
h is a constant =6.63E-34Js.
E=6.63E-34Js × 2.281E14 s-¹              
E=1.512E-19J.
.: the energy of radiation, E=1.512E-19J.

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