Atoms and Nuclei
Results of calculations for four different designs of a fusion reactor using D-D reaction are given below. Which of these is most promising based on Lawson criterion?
Statement- 1:
Energy is released when heavy nuclei undergo fission or light nuclei undergo fusion and
Statement- 2:
For heavy nuclei, binding energy per nucleon increases with increasing $Z$ while for light nuclei it decreases with increasing $Z.$
Then which of the following is true?
A radioactive sample S1 having activity of 5 $\mu$Ci has twice the number of nuclei as another sample S2 which has an activity of 10 $\mu$Ci. The half lives of S1 and S2 can be :
The quantum number n of the state finally populated in He$^+$ ions is :
The wavelength of light emitted in the visible region by He$^+$ ions after collisions with H atoms is
The ratio of the kinetic energy of the $n=2$ electron for the H atom to that of He$^+$ ion is
Assume that the nuclear binding energy per nucleon (B/A) versus mass number (A) is as shown in the figure. Use this plot to choose the correct choice(s) given below.
In the option given below, let E denote the rest mass energy of a nucleus and n a neutron. The correct option is
The largest wavelength in the ultraviolet region of the hydrogen spectrum is 122 nm. The smallest wavelength in the infrared region of the hydrogen spectrum (to the nearest integer) is
energy of proton must be
$ \text { Match the following Columns. } $
| Column I | Column II | ||
|---|---|---|---|
| (A) | Nuclear fusion. | (P) | Converts some matter into energy. |
| (B) | Nuclear fission. | (Q) | Generally possible for nuclei with low atomic number. |
| (C) | $\beta$-decay. | (R) | Generally possible for nuclei with higher atomic number. |
| (D) | Exothermic nuclear reaction. | (S) | Essentially proceeds by weak nuclear forces. |
$ [\mathrm{A} \rightarrow(\mathrm{P}) ; \mathrm{B} \rightarrow(\mathrm{P}, \mathrm{R}) ; \mathrm{C} \rightarrow(\mathbf{P}) ; \mathbf{D} \rightarrow(\mathbf{P}, \mathbf{Q}, \mathbf{R})] . $
$ [\mathrm{A} \rightarrow(\mathrm{P}, \mathrm{Q}) ; \mathrm{B} \rightarrow(\mathrm{P}, \mathrm{R}) ; \mathrm{C} \rightarrow(\mathbf{P}, \mathbf{S}) ; \mathbf{D} \rightarrow(\mathbf{P}, \mathbf{Q}, \mathbf{R})] . $
$ [\mathrm{A} \rightarrow(\mathrm{P}, \mathrm{Q}) ; \mathrm{B} \rightarrow(\mathrm{P}, \mathrm{R}) ; \mathrm{C} \rightarrow( \mathbf{S}) ; \mathbf{D} \rightarrow(\mathbf{P}, \mathbf{Q})] . $
In hydrogen-like atom $(z=11)$, $n$th line of Lyman series has wavelength A equal to the de Broglie's wavelength of electron in the level from which it originated. What is the value of $n$ ?
Explanation:
The $n^{\text {th }}$ line in Lyman series corresponds to transition $(n+1) \rightarrow 1$. The wavelength of this transition is given by
$ \begin{aligned} \frac{1}{\lambda} & =\mathrm{R} z^2\left|\frac{1}{1^2}-\frac{1}{(n+1)^2}\right| \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \mathrm{R} & =1.097 \times 10^7 \mathrm{~m}^{-1} \text { and } z=11 . \end{aligned} $
The angular momemtum in $n^{\text {th }}$ orbit is given by
$ m v r=\frac{n h}{2 \pi} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$
The de-Broglie wavelength of electron in $(n+1)^{\text {th }}$ orbit is
$ \lambda=\frac{h}{m v}=\frac{h r}{m v r}=\frac{2 \pi r}{(n+1)} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$
$ \begin{aligned} &\text { The radius of }(n+1) \text { th orbit is }\\ &r=a_0(n+1)^2 / z \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\end{aligned} $
where, $a_0=0.5269 \times 10^{-10} m$ is Bohr's radius.
Substitute $r$ from equation (iv) into equation (iii) and then substitute into equation (i) to get.
$ \begin{aligned} \lambda & =\frac{2 \pi\left(a_0(n+1)^2\right) / z}{(n+1)} \\ & =\frac{2 \pi a_0(n+1)}{z} \end{aligned} $
From (i)
$ \begin{aligned} & \frac{\frac{1}{\frac{2 \pi a_0(n+1)}{z}}=\mathrm{R} z^2\left|\frac{1}{1^2}-\frac{1}{(n+1)^2}\right|}{\frac{z}{2 \pi a_0(n+1)}=\mathrm{R} z^2\left|\frac{1}{1^2}-\frac{1}{(n+1)^2}\right|} \\ & \frac{z}{2 \pi\left(0.529 \times 10^{-10}\right)(n+1)} \end{aligned} $
$ \begin{aligned} &=1.09 \times 10^7 \times z^2\left[1-\frac{1}{(n+1)^2}\right]\\ &\text { Put } z=11\\ &\begin{aligned} & \frac{11}{2 \pi\left(0.529 \times 10^{-10}\right)(n+1)} \\ & \quad=1.09 \times 10^7(11)^2\left[1-\frac{1}{(n+1)^2}\right] \\ & \frac{1}{2 \pi\left(0.529 \times 10^{-10}\right)(n+1)} \\ & \quad=1.09 \times 10^7(10)\left[1-\frac{1}{(n+1)^2}\right] \\ & \frac{1}{2 \pi\left(0.529 \times 10^{-10}\right)\left(1.09 \times 10^7\right)(11)} \\ & \quad=\left[1-\frac{1}{(n+1)^2}\right][n+1] \end{aligned} \end{aligned} $
$ \begin{aligned} & \frac{1}{2 \pi(0.529)\left(10^{-10}\right)\left(1.09 \times 10^7\right)(11)} \\ & =\frac{(n+1)^2-1}{(n+1)^2} \cdot(n+1) \end{aligned} $
$ \begin{aligned} \frac{1}{2 \pi(0.529)(1.09)(11)\left(10^{-3}\right)} & =\frac{(n+1)^2-1}{(n+1)} \\ \frac{1}{39.852 \times 10^{-3}} & =\frac{(n+1)^2-1}{(n+1)} \\ \frac{1000}{39.852} & =\frac{(n+1)^2-1}{(n+1)} \\ 25.09 & =\frac{(n+1)^2-1}{n+1} \\ \Rightarrow \quad 25 n+25=n^2+2 n & +1-1 \\ & \quad \text { (approximate) } \end{aligned} $
$ \begin{aligned} & \quad \begin{aligned} 25 n & +25=n^2+2 n \\ n^2-23 n & -25=0 \\ n & =\frac{-(-23) \pm \sqrt{(-23)^2-4.1 .(-25)}}{2.1} \\ & =\frac{23 \pm \sqrt{529+100}}{2}=\frac{23 \pm \sqrt{629}}{2} \\ n & =\frac{23+\sqrt{629}}{2}=\frac{23-\sqrt{629}}{2} \\ n & =\frac{23+25}{2}=\frac{23-25}{2} \\ \quad n & =24, n=1 \\\therefore n & =24 \end{aligned} \end{aligned} $
Highly energetic electrons are bombarded on a target of an element containing 30 neutrons. The ratio of radii of nucleus to that of Helium nucleus is $(14)^{\frac{1}{3}}$. Find
(A) Atomic number of the nucleus;
(B) the frequency of $\mathrm{K}_{\alpha}$ line of the X-ray produced.
$\left(\mathrm{R}=1.1 \times 10^{7} \mathrm{~m}^{-1}\right.$ and $\left.c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)$
given that the repulsive potential energy between the two nuclei is $ \sim 7.7 \times {10^{ - 14}}J$, the temperature at which the gases must be heated to initiate the reaction is nearly
[ Boltzmann's Constant $k = 1.38 \times {10^{ - 23}}\,J/K$ ]
Then $Z$ of the resulting nucleus is
$\eqalign{ & \left( i \right)\,\,\,\,\,\,\,electrons\,\,\,\,\,\,\,\,\,\,\,\,\left( {ii} \right)\,\,\,\,\,\,\,protons \cr & \left( {iii} \right)\,\,\,H{e^{2 + }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {iv} \right)\,\,\,\,\,\,\,neutrons \cr} $
The emission at instant can be