Let v_{n} and E_{n} be the respective speed a | vedclass

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

Question

(A-D) According to Bohr's model for a hydrogen atom:$v_{n} \propto \frac{1}{n}$$E_{n} \propto \frac{1}{n^{2}}$$r_{n} \propto n^{2}$For option $A$: $\f

Vedclass · Accepted Answer

plot of $\ln \left(\frac{r_{n} E_{1}}{E_{n} r_{1}}\right)$ as a function of $\ln (n)$ is a straight line of slope $4$

Vedclass · Answer

(A-D) According to Bohr's model for a hydrogen atom:$v_{n} \propto \frac{1}{n}$$E_{n} \propto \frac{1}{n^{2}}$$r_{n} \propto n^{2}$For option $A$: $\frac{E_{n} r_{n}}{E_{1} r_{1}} \propto \frac{(1/n^{2}) \cdot n^{2}}{1} = 1$. This is a constant, so the slope is $0$.For option $B$: $\frac{r_{n} v_{n}}{r_{1} v_{1}} \propto \frac{n^{2} \cdot (1/n)}{1} = n$. This is a straight line with slope $1$.For option $C$: $\frac{r_{n}}{r_{1}} = n^{2}$. Taking natural log on both sides: $\ln \left(\frac{r_{n}}{r_{1}}\right) = 2 \ln(n)$. This is a straight line with slope $2$.For option $D$: $\frac{r_{n}}{E_{n}} \propto \frac{n^{2}}{1/n^{2}} = n^{4}$. Thus, $\frac{r_{n} E_{1}}{E_{n} r_{1}} = n^{4}$. Taking natural log: $\ln \left(\frac{r_{n} E_{1}}{E_{n} r_{1}}\right) = 4 \ln(n)$. This is a straight line with slope $4$.All options $A, B, C,$ and $D$ are mathematically correct based on Bohr's model.

Let $v_{n}$ and $E_{n}$ be the respective speed and energy of an electron in the $n$th orbit of radius $r_{n}$, in a hydrogen atom, as predicted by Bohr's model. Then:

Similar Questions

The magnetic moment of an electron due to its orbital motion is proportional to ($n$ = principal quantum number).

Let $A_n$ be the area enclosed by the $n^{th}$ orbit in a hydrogen atom. The graph of $\ln (A_n/A_1)$ against $\ln (n)$:

If an electron is revolving in its Bohr orbit having Bohr radius of $0.529 Å$,then the radius of the third orbit is

The ratio of the speed of an electron in the ground state of the Bohr's first orbit of a hydrogen atom to the velocity of light in air is:

Excitation energy of a hydrogen-like ion in its first excitation state is $40.8 \, eV$. Energy needed to remove the electron from the ion in the ground state is ........ $eV$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module