IIT JEE and CBSE Board Numerical

**1.$$ An alpha particle (helium nucleus) of energy 5.5 MeV is scattered through 180° by a gold nucleus. The distance of the closest approach is: Shortcut: The distance of closest approach is given by: $$b = \frac{2Z_1Z_2e^2}{K}$$

Where:

• (Z_1) and (Z_2) are the atomic numbers of the two nuclei.
• (e) is the elementary charge.
• (K) is the kinetic energy of the alpha particle.

Substituting the given values, we get:

$$b = \frac{2(2)(79)(1.6 \times 10^{−19} \text{ C})^2}{5.5 \times 10^6 \text{ eV}}$$ $$b = 30 \times 10^{−15} \text{ m}$$ $$b = 30 \text{ fm}$$

**2.$$ A beam of electrons (charge -e, mass m) is incident on a thin gold foil. The electrons are scattered through an angle of 120°. The de Broglie wavelength of the electrons is:

Shortcut: The de Broglie wavelength of an electron is given by: $$\lambda = \frac{h}{p}$$

Where:

• (h) is the Planck constant.
• (p) is the momentum of the electron.

The momentum of an electron is given by: $$p = mv$$

Where:

• (m) is the mass of the electron.
• (v) is the velocity of the electron.

The velocity of an electron can be found using the kinetic energy of the electron: $$K = \frac{1}{2}mv^2$$

Where:

• (K) is the kinetic energy of the electron.

Substituting the given values, we get: $$5.5 \times 10^6 \text{ eV} = \frac{1}{2}(9.11 \times 10^{−31} \text{ kg})v^2$$ $$v = 2.09 \times 10^7 \text{ m/s}$$

The de Broglie wavelength of the electrons is then: $$\lambda = \frac{6.626 \times 10^{−34} \text{ J s}}{9.11 \times 10^{−31} \text{ kg} \times 2.09 \times 10^7 \text{ m/s}}$$ $$\lambda = 0.024 \text{ nm}$$ $$0.024 \text{ nm}$$ 3. A hydrogen atom is in the ground state. The electron is excited to the n = 3 state by absorbing a photon of light. The wavelength of the photon is:

Shortcut: The energy of a photon is given by: $$E = hf$$

Where:

• (E) is the energy of the photon.
• (h) is the Planck constant.
• (f) is the frequency of the photon.

The frequency of a photon can be found using the wavelength of the photon: $$c = f\lambda$$

Where:

• (c) is the speed of light.

The energy of the photon is also equal to the difference in energy between the two states: $$E = E_3 - E_1$$

Where:

• (E_3) is the energy of the n = 3 state.
• (E_1) is the energy of the ground state.

The energies of the states of a hydrogen atom are given by: $$E_n = \frac{-13.6 \text{ eV}}{n^2}$$

Where:

• (n) is the principal quantum number.

Substituting the given values, we get: $$E = -13.6 \text{ eV} \left(\frac{1}{3^2} - \frac{1}{1^2}\right)$$ $$E = 10.2 \text{ eV}$$

The wavelength of the photon is then: $$\lambda = \frac{hc}{E}$$

$$\lambda=\frac{(6.626\times10^{-34})(3\times10^8)}{10.2\times1.6\times10^{-19}}$$

$$=656\times10^{-9}=656\space nm$$

$$=656\space nm$$ 4. A helium atom has two electrons in the n = 1 state and two electrons in the n = 2 state. The energy of the helium atom is:

Shortcut: The energy of a helium atom can be calculated by adding the energies of the two electrons in the n = 1 state and the two electrons in the n = 2 state. The energy of an electron in the n = 1 or 2 state is given by the formula: $$E_n=-\frac{13.6}{n^2}\space eV$$ The total energy for the n= 1 and n = 2 states becomes: $$E_1=2\times\left ( \frac{-13.6}{1^2} \right )=-27.2\space eV$$ $$E_2=2\times \left ( \frac{-13.6}{2^2} \right )=-6.8\space eV$$ Adding both the energy, we get,

$$E=27.2-6.8=-19.4\space eV$$

Since helium has 2 electrons so, the energy will be doubled $$E=2\times(-19.4)=-38.8\space eV$$ Therefore, the energy of the helium atom is $$-38.8 \space eV$$.

5. A lithium atom has three electrons in the n = 1 state and one electron in the n = 2 state. The energy of the lithium atom is:

Shortcut: The energy of a lithium atom can be calculated by adding the energies of the three electrons in the n = 1 state and the one electron in the n = 2 state:

$$E=-3\times\left ( \frac{-13.6}{1^2} \right )- \left ( \frac{-13.6}{2^2} \right )$$ $$E=-3\times(-13.6)-(-6.8)$$ $$E=32.8\space eV$$

Therefore, the energy of the lithium atom is $$-32.8 \space eV$$.

6. A beryllium atom has four electrons in the n = 1 state and two electrons in the n = 2 state. The energy of the beryllium atom is:

Shortcut: The energy of a beryllium atom can be calculated by adding the energies of the four electrons in the n = 1 state and the two electrons in the n = 2 state:

$$E=-4\times\left ( \frac{-13.6}{1^2} \right )- 2\times\left ( \frac{-13.6}{2^2} \right )$$ $$E=-4\times(-13.6)-2\times(-6.8)$$ $$E=54.4+13.6=68\space eV$$

Therefore, the energy of the beryllium atom is $$-68 \space eV$$.

7. A boron atom has five electrons in the n = 1 state and two electrons in the n = 2 state. The energy of the boron atom is:

Shortcut: The energy of a boron atom can be calculated by adding the energies of the five electrons in the n = 1 state and the two electrons in the n = 2 state:

$$E=-5\times\left ( \frac{-13.6}{1^2} \right )- 2\times\left ( \frac{-13.6}{2^2} \right )$$ $$E=-5\times(-13.6)-2\times(-6.8)$$ $$E=68+13.6=81.6 \space eV$$

Therefore, the energy of the boron atom is $$-81.6\space eV$$.

8. A carbon atom has six electrons in the n = 1 state and two electrons in the n = 2 state. The energy of the carbon atom is:

Shortcut: The energy of a carbon atom can be calculated by adding the energies of the six electrons in the n = 1 state and the two electrons in the n = 2 state:

$$E=6\times\left ( \frac{-13.6}{1^2} \right )+ 2\times\left ( \frac{-13.6}{2^2} \right )$$ $$E=6\times(-13.6)+2\times(-6.8)$$ $$E=-81.6-13.6=-95.2 \space eV$$

Therefore, the energy of the carbon atom is $$-95.2 \space eV$$.

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