Answered: I was trying to comprehend this problem… | bartleby

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Hi! I was trying to comprehend this problem (see attached photo). Bohr's atomic model states that the total energy of an electron in any orbit is calculated with the formula: E_n=-(2π²me⁴Z²k²/n²h²(4πε₀)²). However, in the attached photo, the part (4πε₀)² is missing in the denominator. May I know what happened here?

Radius of the eleebon's dbit in given by-
%3D
here,
paimcipal quantum number
Planck's Can tant
ニe
h =
m = ma ss q electron
K = पाट.
2 = atomic number
e= electronic Charoge.
dbit
any
TO tal enersy of eleetoun in
h given by-
En =
- an me4z?K?
Pathng the value
%3D
2スK
ス=2
tor He
%3D
When - 30 mm =30x109 m
e'k
30xi09
and For , =207m =
then eneny
E =
%3D
X16 9 m 20י .
,enersy
E2 =
720x109 J
Now,
enersy of the photon = enersy
differinee between two dbi
E, -E = (-
3.84x10 9 S
2.4 ev
e2K
30X109
e?K
e= 1-6 x 10'9 C
%3D
%3D
| 1 ev = l-6 X 10 19J

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