Question

Q.Using Rutherford model of the atom, derive the expression for the total energy of the electron in hydrogen atom. What is the significance of total negative energy possessed by the electron?​

Answer 1

ANSWER-;

According to Rutherford's model, we have

$\frac{mv {}^{2} }{r} = \frac{1}{ \:4πϵ } \frac{ze {}^{2} }{r {}^{2} }$

$⟹\frac{1}{ \:4πϵ } \frac{ze {}^{2} }{r}$

Therefore,

Therefore,Total Energy =P.E+K.E

$T.E =- \frac{1}{ \:4πϵ} \frac{ze {}^{2} }{r {}^{} } + \frac{1}{2} mv {}^{2}$

$= \frac{1}{2} \frac{1}{ 4πϵ } \frac{ze {}^{2} }{r}$

$= - \frac{1}{ 8πϵ } \frac{ze {}^{2} }{r}$

Energy is negative implies that the electron- nucleus is a bound or attractive system.

Answer 2

Answer:

According to Rutherford's model, we have

$\frac{ {mv}^{2} }{r} = \frac{1}{4\pi ϵ </p><p>o} \frac{ {ze}^{2} }{ {r}^{2} }$

${mv}^{2} = \frac{1}{4\pi ϵ </p><p>o} \frac{ {ze}^{2} }{r}$

Therefore,

total energy = P.E + K.E

T.E =

$- \frac{1}{4\pi ϵ o} \frac{ {ze}^{2} }{r} + \frac{1 }{2} {mv}^{2}$

$- \frac{1}{2} .\frac{1}{4\pi ϵ </p><p>o} \frac{ {ze}^{2} }{r}$

$= - . \frac{1}{8\pi ϵ </p><p>o} \frac{ {ze}^{2} }{r}$

Energy is negative implies that the electron - nucleas is a bound or attractive system.