Atoms Test - 29

The energy emitted by a source is in the form of gamma radiation which is called photons. It is extremely high-frequency electromagnetic radiation.

The radius of first orbit of hydrogen atom is:
$$r = \dfrac{n^2h^2\epsilon_0}{\pi m e^2}$$ ........................(since, n = 1)

The principal quantum number n, belongs to the principal shell (K,L,M, ....) which is occupied by an electron and can be any positive integer (n = 1, 2, 3, 4, ........). Its location is further narrowed down by the angular momentum quantum number l, corresponding to the subshell and its general shape. The value of l is dependent on the value of n (l = 0, 1, 2, .........., n-1). The values of l correspond to specific subshells i.e. l = 0 for s; l = 1 for p; l = 2 for d; l = 3 for f. 
The magnetic quantum number($$m_l$$), further divides these subshells into orbitals and gives orientation of these orbitals in space. The number of orbitals can be determined from the angular momentum quantum number l and the maximum possible values of $$m_l$$ can be determined from number of orbitals given by, (2l + 1).
Also, $$m_l$$ is the projection of l in field direction. For angle of projection $$\theta$$ it is given by
$$m_l = l \cos \theta$$.
Since, $$m_l$$ is an integer and $$\cos \theta$$ never exceed unity hence, the possible values of $$m_l$$ are
$$m_l = -l, ........, 0, ..........., +l$$
Hence, for a hydrogen atom is in the d-state, $$l = 2$$, 
Number of orbitals are,
$$(2l + 1) = (2(2) + 1) = 5$$
Therefore, $$m_l = -2, -1, 0, 1, 2$$

From the $$n^{th}$$ state, the electron may go to $$(n - 1)^{th}$$ state, ...., 2nd state or 1st state. So there are (n 1) possible transitions starting from the $$n^{th}$$ state. The atoms reaching $$(n - 1)^{th}$$ state may make (n 2) different transitions. The atoms reaching $$(n - 2)^{th}$$ state may make (n 3) different transitions. Similarly for other lower states. During each transition a photon with energy $$h\nu$$ and wavelength $$\lambda$$ is emitted out. Hence, the total number of possible transitions is equal to the number of photons emitted. The total number of possible transitions is
$$(n-1), (n-2), (n-3), ................, 3, 2, 1 = \dfrac{n(n-1)}{2}$$

When a hydrogen atom emits a photon of energy 12.1 eV, i.e. the energy difference between two levels hence,
$$E_{n_2} - E_{n_1} = 12.1$$
$$E_{n_2} = 12.1 + E_{n_1}$$
$$E_{n_2} = 12.1 + (-13.6)$$
$$E_{n_2} = -1.5 eV$$

The principle quantum number $$n = 4$$, represents the fourth orbit. A subshell is the set of states defined by a common azimuthal quantum number l, within a shell. The values $$l = 0, 1, 2, 3$$ correspond to the s, p, d, and f shells, respectively. Hence, $$n = 4$$, $$l = 1$$ represents 4p subshell.

The radius of an orbit of revolution of electron is:  $$r = \dfrac{n^2h^2\epsilon_0}{\pi m e^2}$$ 
The velocity of revolving electron is:  $$v = \dfrac{e^2}{2nh\epsilon_0}$$
Hence, $$r \propto n^2$$ and $$v \propto \dfrac{1}{n}$$
Therefore, decreasing principal quantum number $$n, r$$ will decrease and $$v$$ will increase.

We know that the first shell consists of only s orbital i.e. 1 subshell and can accommodate 2 electrons. The second shell consists of two orbitals namely s and p and it can accommodated 2 electrons in the s orbital and 6 electrons in the p orbital i.e. a total of 8 electrons in the second shell.
Likewise, the d subshell accommodates 10 electrons and f subshell accommodates 14 electrons.
Thus we see that the maximum electrons in any subshell is given using $$2(2l+1).$$
For $$l=0$$ we have $$n=2$$
For $$l=1$$ we have $$n=6$$
For $$l=2$$ we have $$n=10$$
For $$l=3$$ we have $$n=14$$
($$l$$ is the azimuthal quantum number)

Values for the quantum number $$m_{l}$$ are -1,0,+1
Hence, option A.

Spin of an electron, neutron and proton is $$\dfrac{h}{2}$$.