The difference between the n^{text{th}} and ( | vedclass

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

Question

(D) The formula for the $n^{	ext{th}}$ Bohr radius is $r_n = a_0 n^2$,where $a_0$ is the Bohr radius constant.Given the condition: $r_{n+1} - r_n = r

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) The formula for the $n^{	ext{th}}$ Bohr radius is $r_n = a_0 n^2$,where $a_0$ is the Bohr radius constant.Given the condition: $r_{n+1} - r_n = r_{n-1}$.Substituting the formula: $a_0(n+1)^2 - a_0 n^2 = a_0(n-1)^2$.Dividing by $a_0$: $(n+1)^2 - n^2 = (n-1)^2$.Expanding the terms: $(n^2 + 2n + 1) - n^2 = n^2 - 2n + 1$.Simplifying: $2n + 1 = n^2 - 2n + 1$.Rearranging the equation: $n^2 - 4n = 0$.Factoring: $n(n - 4) = 0$.Since $n$ must be a positive integer for Bohr orbits,$n = 4$.

The difference between the $n^{\text{th}}$ and $(n+1)^{\text{th}}$ Bohr's radius of a hydrogen atom is equal to the $(n-1)^{\text{th}}$ Bohr's radius. The value of $n$ is:

Similar Questions

What type of polymer is natural rubber?

The products formed when diethyl ether is reacted with cold $HI$ are

In a slide caliper, \((m+1)\) number of vernier divisions is equal to \(m\) number of smallest main scale divisions. If \(d\) unit is the magnitude of the smallest main scale division, then the magnitude of the vernicr constant is

The distance between the pair of lines $x^2+2 \sqrt{2} x y+2 y^2+4 x+4 \sqrt{2} y+1=0$ is:

$A$ metal rod of length $2 \text{ m}$ has cross-sectional areas $2A$ and $A$ as shown in the figure. The ends are maintained at temperatures $100^{\circ}\text{C}$ and $70^{\circ}\text{C}$. The temperature at the middle point $C$ is...... $^{\circ}\text{C}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module