accelerated potential = 1 keV
The kinetic energy of the electron = The energy due to accelerating potential.
\(\frac{1}{2} m v^{2}=e V\)
mv² = 2eV
\( m^{2} v^{2}=2 m e V \Rightarrow(m v)^{2}=2 m e V \)
\(\Rightarrow \mathrm{mv}=\sqrt{2 \mathrm{meV}}\)
de-Broglie wavelength \( \lambda=\frac{\mathrm{h}}{\mathrm{mv}} \)
\(\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}}\)
m = Mass of the electron = 9.1 x 10^-31 kg
h - Planck'sconstant = 6.626 x 10^-31Js
1 eV = 1.6 x 10^-19J
\( \lambda=\frac{6.626 \times 10^{-34} \mathrm{Js}}{\sqrt{2 \times 9.1 \times 10^{-31} \mathrm{~kg} \times 1 \mathrm{keV}}} \)
\(\lambda=\frac{6.626 \times 10^{-34} \mathrm{JS}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1 \times 10^{3} \times 1.6 \times 10^{-19} \mathrm{kgJ}}} \)
\({\left[\because \frac{\mathrm{Js}}{\sqrt{\mathrm{Jkg}}}=\mathrm{J}^{1 / 2} \mathrm{~kg}^{-1 / 2} \cdot \mathrm{s}=\left(\mathrm{kgm}^{2} \mathrm{~s}^{-2}\right)^{1 / 2} \mathrm{~kg}^{-1 / 2} \cdot \mathrm{s}=\mathrm{m}\right]}\)
= 3.88 x 10^-11 m

Given \(\triangle\)v = 5.7 x 10⁵ ms^-1. \(\triangle\)x = ?
According to Heisenbergs uncertainty principle \(\Delta x \cdot \Delta p \geq \frac{\mathrm{h}}{4 \pi}\)
\( \frac{\mathrm{h}}{4 \pi}=\frac{6.626 \times 10^{-34}}{4 \times 3.14} \mathrm{kgm}^{2} \mathrm{~s}^{-1}=5.28 \times 10^{-35} \)
\(\Delta x \cdot \Delta \mathrm{p} \geq 5.28 \times 10^{-35} \)
\(\Delta x . \mathrm{m} \Delta \mathrm{v} \geq 5.28 \times 10^{-35} \)
\(\Rightarrow \Delta x \geq \frac{5.28 \times 10^{-35} \mathrm{kgm}^{2} \mathrm{~s}^{-1}}{9.1 \times 10^{-31} \mathrm{~kg} \times 5.7 \times 10^{5} \mathrm{~ms}^{-1}} \Rightarrow \Delta x \geq 1.017 \times 10^{-10} \mathrm{~m}\)

The variation of the probability of locating the electron on a sphere with nucleus at its centre depends on the azimuthal quantum number of the orbital in which the electron is present.

For 1s orbital, l = 0, m = 0,\(f(\theta)=\frac{1}{\sqrt 2}\)and \(g(\varphi)=\frac{1}{\sqrt 2\pi}\) Therefore, the angular distribution function is equal to \(\frac{1}{2\sqrt \pi }\) 
i.e. it is independent of the angle \(\theta\) and \(\varphi\). Hence, the probability of finding the electron is independent of the direction from the nucleus. So, the shape of the s orbital is spherical.

For p orbitals, l = 1 and the corresponding m values are -1, 0 and +1. The angular distribution functions are quite complex and are not discussed here, The shape of the p orbital is shown in Figure (b). The three different m values indicates that there are three different orientations possible for p orbitals. These orbitals are designated as P_x, P_y and P_z and the angular distribution for these orbitals shows that the lobes are along the x, y arid z axis respectively. As seen in the Figure the 2p orbitals have one nodal plane

For 'd' orbital l = 2 and the corresponding m values are -2, -1, 0, +1, +2. The shape of the d orbital looks like a 'clover leaf'.
The five m values give rise to five d orbitals namely d_xy, d_yz, d_zx, d_x²-y² and d_z². The 3d orbitals contain two nodal planes.

For 'f' orbital, l = 3 and the m values are -3, -2,-1,0, +1, +2, +3 corresponding to seven f orbitals. f_z³, f_xz²,f_yz²,f_xyz,f_{z(x² - y²)},f_{x(x² - 3y²)},f_{y(3x² - y²)} which are shown in Figure. There are 3 nodal planes in the f-orbitals.

(i) Magnetic quantum number
1. It is denoted by the letter 'm_l'. It takes integral values ranging from -I to +1 through 0. i.e. if l = 1; m = -1, 0 and +1.
2. The Zeeman Effect (the splitting of spectral lines in a magnetic field) provides the experimental justification for this quantum number.
3. The magnitude of the angular momentum is determined by the quantum number l while its direction is given by magnetic quantum number.
(ii) Spin quantum number
1. The spin quantum number represents the spin of the electron and is denoted by the letter 'm_s'.
2. The electron in an atom revolves not only around the nucleus but also spins. It is usual to write this as electron spins about its own axis either in a clockwise direction or in anti-clockwise direction.
3. Corresponding to the clockwise and anti-clockwise spinning of the electron, maximum two values are possible for this quantum number.
4. The values of 'm_s' is equal to \(-\frac{1}{2}\) and  \(+\frac{1}{2}.\)