When an electron makes a transition from (n+1 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The Rydberg formula is given by $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$,where $n_1 = n$ and $n_2 = n+1$.Subst

Vedclass · Accepted Answer

$v = \frac{2cR Z^2}{n^3}$

Vedclass · Answer

(A) The Rydberg formula is given by $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$,where $n_1 = n$ and $n_2 = n+1$.Substituting the values: $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} \right)$.Simplifying the expression: $\frac{1}{\lambda} = R Z^2 \left( \frac{(n+1)^2 - n^2}{n^2(n+1)^2} \right) = R Z^2 \left( \frac{2n+1}{n^2(n+1)^2} \right)$.Given the condition $n >> 1$,we can approximate $2n+1 \approx 2n$ and $(n+1)^2 \approx n^2$.Thus,$\frac{1}{\lambda} \approx R Z^2 \left( \frac{2n}{n^2 \cdot n^2} \right) = \frac{2R Z^2}{n^3}$.Since $v = \frac{c}{\lambda}$,we have $v = \frac{2cR Z^2}{n^3}$.

When an electron makes a transition from $(n+1)$ state to $n^{th}$ state,the frequency of emitted radiation is related to $n$ according to $(n >> 1)$:

Similar Questions

What is the maximum wavelength line in the Lyman series of $He^{+}$ ion?

Find the frequency of light that corresponds to photons of energy $5.0 \times 10^{-5} \ erg$.

If $P.E.$ of $3^{rd}$ shell is $-20 \ eV/atom$,then find the $I.P.$ of the $3^{rd}$ shell of that element.

The electron present in an excited state of $He^{+}$ ion shows separation (ionization) on collision with an electron having $2.5 \ eV$ energy. What will be the minimum value of $n$ for this electron of $He^{+}$ ion?

The energy of an electron revolving in the $n^{th}$ Bohr's orbit of an atom is given by the expression

Vedclass Test Series

Exam Paper Generator

Online Exam Module