The Correct Answer is Option (C)

The statement about the binding energy per nucleon is true, and is known as the semi-empirical mass formula. According to this formula, the binding energy per nucleon for nuclei in the range of mass numbers 30 to 170 is nearly constant, with a maximum value around mass number 60.

The statement about nuclear force being short ranged is also true. The strong nuclear force that binds nucleons together is a short-range force that acts only over distances of a few femtometers.

Therefore, both Assertion A and Reason R are true, and Reason R provides a valid explanation for Assertion A. The correct answer is:

So, Both Assertion A and Reason R are true, and Reason R is the correct explanation for Assertion A.

The Correct Answer is Option (A)

We are given that the half-life of A is the same as the average life of B. The relationship between half-life ($T_{1/2}$) and the decay constant ($\lambda$) is:

$T_{1/2}=\frac{\ln 2}{\lambda }$

For the average life ($\tau$), the relationship with the decay constant is:

$\tau =\frac{1}{\lambda }$

According to the given information, the half-life of A is equal to the average life of B:

$T_{1/2(A)}={\tau }_B$

Now, we can substitute the relationships for half-life and average life:

$\frac{\ln 2}{{\lambda }_A}=\frac{1}{{\lambda }_B}$

To find the correct relationship between ${\lambda }_A$ and >${\lambda }_B$, we can rearrange the equation:

${\lambda }_A={\lambda }_B\ln 2$

The Correct Answer is Option (A)

According to Bohr's model of the hydrogen atom, the angular momentum of an electron in an orbit is an integral multiple of Planck's constant divided by $2\pi$ (or $h/2\pi$, where $h$ is the Planck's constant). This can be expressed as:

$L=n\frac{h}{2\pi }$

where $n$ is the principal quantum number or the orbit number.

So, for the first orbit ($n=1$), the angular momentum $L_1$ is:

$L_1=1\times \frac{h}{2\pi }=\frac{h}{2\pi }$

And for the second orbit ($n=2$), the angular momentum $L_2$is:

$L_2=2\times \frac{h}{2\pi }=\frac{h}{\pi }$

The change in angular momentum when moving from the first to the second orbit is the difference between $L_2$ and $L_1$:

$\mathrm{\Delta }L=L_2-L_1=\frac{h}{\pi }-\frac{h}{2\pi }=\frac{h}{2\pi }$

Since $\frac{h}{2\pi }$ is equal to the initial angular momentum $L_1$, the change in angular momentum when the electron moves to the second orbit is $L$.

Therefore, the correct answer is $L$.

The Crrect Answer is

Let the electron jumps to $n^{}$ orbit so

So, Angular momentum $\begin{array}{rl}& 12.75=13.6[\frac{1}{1^2}-\frac{1}{n^2}]\\ \\ & \Rightarrow n=4\\ \\ & L=\frac{nh}{2\pi }=\frac{2h}{\pi }\end{array}$

$\begin{array}{rl}\text{Angular momentum}=\frac{2}{\pi }& \times 4.14\times {10}^{-15}\\ \\ & =\frac{828\times {10}^{-17}}{\pi }\mathrm{eVs}\end{array}$

The Crrect Answer is

$\frac{1}{\lambda }={\mathrm{RZ}}^2[\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}]$

$\begin{array}{rl}& \frac{1}{{\lambda }_1}={\mathrm{RZ}}^2[\frac{1}{1^2}-\frac{1}{3^2}]=\frac{8}{9}{\mathrm{RZ}}^2........(1)\\ \\ & \frac{1}{{\lambda }_2}={\mathrm{RZ}}^2[\frac{1}{1^2}-\frac{1}{2^2}]=\frac{3}{4}{\mathrm{RZ}}^2........(2)\end{array}$

$\begin{array}{c}1/2\Rightarrow \frac{{\lambda }_2}{{\lambda }_1}=\frac{8}{9}\times \frac{4}{3}=\frac{32}{27}\\ \\ \frac{{\lambda }_1}{{\lambda }_2}=\frac{27}{32}\end{array}$