How many revolutions an election will make in 2nd shell of He+ in one second
Answers
Answer: 2.2429 × 10¹⁴
Step-by-step explanation:
We know that,
According to bohr's atomic model,
Radius of the nth orbit is given by,
r_n=\frac{h^2}{4\pi me^2}\times\frac{n^2}{Z}r
n
=
4πme
2
h
2
×
Z
n
2
Where,
h(Plank's Constant) = 6.62 × 10⁻²⁷ arg-second
m(Mass of electron) = 9.109 × 10⁻²⁸ gram
e(Charge on electron) = 4.808 × 10⁻¹⁰ esu
Putting these values in above equation, We get a simplified equation,
\begin{lgathered}r_n=0.529\times10^{-8}\times\frac{n^2}{Z}\;cm\\\;\\r_n=0.529\times10^{-10}\times\frac{n^2}{Z}\;m\end{lgathered}
r
n
=0.529×10
−8
×
Z
n
2
cm
r
n
=0.529×10
−10
×
Z
n
2
m
For Velocity of electron in nth orbital,
V_n=\frac{2\pi e^2}{h}\times\frac{Z}{n}V
n
=
h
2πe
2
×
n
Z
On Simplifying the equation
V_n=2.18\times10^6\times\frac{Z}{n}\;\;m/sV
n
=2.18×10
6
×
n
Z
m/s
Now, Revolution per second means Frequency of electron
\begin{lgathered}f=\frac{V_n}{2\pi r_n}\\\;\\f=\frac{2.18\times10^6\times\frac{Z}{n}}{2\pi\times0.529\times10^{-10}\times\frac{n^2}{Z}}\\\;\\f=\frac{0.6559\times10^{16}}{n^3}\end{lgathered}
f=
2πr
n
V
n
f=
2π×0.529×10
−10
×
Z
n
2
2.18×10
6
×
n
Z
f=
n
3
0.6559×10
16
For Hydrogen atom , Z = 1 and 3rd orbital n = 3
\begin{lgathered}f=\frac{0.6559\times10^{16}}{3^3}\\\;\\f=0.02429\times10^{16}\\\;\\f=2.2429\times10^{14}\;\text{revolutions per second}\end{lgathered}
f=
3
3
0.6559×10
16
f=0.02429×10
16
f=2.2429×10
14
revolutions per second