a) State and explain Bohr's postulates of an atomic model. [2] b) i) Calculate the de Broglie wavelength of a neutron having kinetic energy 150eV . [2] ii) A proton and an electron have same K.E.…

a) Bohr’s Postulates of the Atomic Model

Niels Bohr proposed five key postulates to explain the stability of atoms and the emission/absorption of spectral lines. These are:

1. Stationary Orbits (Quantization of Angular Momentum)

   • Electrons revolve around the nucleus in fixed, circular orbits called stationary orbits without radiating energy.
   • Only certain orbits are allowed, where the angular momentum $L$ of an electron is quantized: $$L=mvr=n\frac{h}{2\pi }(n=1,2,3,\dots )$$ where $m$ = electron mass, $v$ = orbital velocity, $r$ = radius, $h$ = Planck’s constant, and $n$ = principal quantum number.

2. Energy Levels (Discrete Energy States)

   • Each stationary orbit corresponds to a specific energy level $E_n$. The energy of the $n$th orbit is given by: $$E_n=-\frac{13.6}{n^2}\text{ eV}$$ (for hydrogen; the negative sign indicates bound states).

3. Radiation Emission/Absorption

   • An electron emits or absorbs energy only when it jumps between orbits. The energy of the emitted/absorbed photon is: $$\Delta E=E_f-E_i=h\nu =h\frac{c}{\lambda }$$ where $\nu$ is frequency and $\lambda$ is wavelength.

4. Electron’s Motion in Orbits

   • The centripetal force for the electron’s circular motion is provided by the electrostatic attraction (Coulomb’s law): $$\frac{m{v}^2}{r}=\frac{e^2}{4\pi {ϵ}_0{r}^2}$$ This relates the electron’s velocity and radius to the orbit number $n$.

5. Ground State and Excitation

   • The lowest energy orbit ($n=1$) is the ground state. Higher orbits ($n>1$) are excited states. Electrons can be excited to higher levels by absorbing energy (e.g., heat, light) and return to lower levels by emitting photons.

Significance: Bohr’s model successfully explained the hydrogen spectrum (Balmer, Lyman series) and introduced the concept of quantized energy levels, laying the foundation for quantum mechanics.

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b) i) De Broglie Wavelength of a Neutron

Given:

• Kinetic energy of neutron, $K=150\text{ eV}$.
• Mass of neutron, $m_n=1.675\times 10^{-27}\text{ kg}$.
• Planck’s constant, $h=6.626\times 10^{-34}\text{ J s}$.
• $1\text{ eV}=1.602\times 10^{-19}\text{ J}$.

Step 1: Convert kinetic energy to joules. $$K=150\text{ eV}\times 1.602\times 10^{-19}\text{ J/eV}=2.403\times 10^{-17}\text{ J}.$$

Step 2: Relate kinetic energy to momentum. For non-relativistic particles ($K\ll m{c}^2$): $$K=\frac{p^2}{2m}⟹p=\sqrt{2mK}.$$ Substitute values: $$p=\sqrt{2\times 1.675\times 10^{-27}\times 2.403\times 10^{-17}}=3.13\times 10^{-23}\text{ kg m/s}.$$

Step 3: Calculate de Broglie wavelength. $$\lambda =\frac{h}{p}=\frac{6.626\times 10^{-34}}{3.13\times 10^{-23}}=2.12\times 10^{-11}\text{ m}=0.0212\text{ nm}.$$

Final Answer: The de Broglie wavelength of the neutron is $0.0212\text{ nm}$.

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b) ii) Comparison of De Broglie Wavelengths

Given:

• A proton and an electron have the same kinetic energy $K$.

Key Relationship: De Broglie wavelength $\lambda =\frac{h}{p}$, and momentum $p=\sqrt{2mK}$. Thus: $$\lambda =\frac{h}{\sqrt{2mK}}.$$ For the same $K$, $\lambda \propto \frac{1}{\sqrt{m}}$.

Masses:

• Electron: $m_e=9.11\times 10^{-31}\text{ kg}$.
• Proton: $m_p=1.67\times 10^{-27}\text{ kg}$.

Comparison: Since $m_p\gg m_e$, $\sqrt{m_p}\gg \sqrt{m_e}$, so: $${\lambda }_e\gg {\lambda }_p.$$

Conclusion: The electron has the greater de Broglie wavelength when both have the same kinetic energy.