Question

The expression for the speed of electron \( v \) in terms of radius of the orbit (\( r \)) and physical constant \[ K = \frac{1}{4 \pi \varepsilon_0} \] is:

• A. \( \frac{Ke^2}{mr} \)
• B. \( \frac{Ke^2}{mr^2} \)
• C. \( \sqrt{\frac{Ke^2}{mr}} \)
• D. \( \sqrt{\frac{Ke^2}{mr^2}} \)
• A. \( \frac{Ke^2}{r} \)
• B. \( -\frac{Ke^2}{2r} \)
• C. \( \frac{Ke^2}{2r} \)
• D. \( \frac{3}{2} \cdot \frac{Ke^2}{r} \)
• A. \( 2.48, -2.48 \)
• B. \( -1.24, 1.24 \)
• C. \( -2.48, 2.48 \)
• D. \( 1.24, -1.24 \)
• A. \( n \)
• B. \( \frac{1}{n} \)
• C. \( \frac{1}{n^2} \)
• D. \( \frac{1}{n^3} \)
• A. \(0.53\ \text{\AA}\)
• B. \(1.06\ \text{\AA}\)
• C. \(1.59\ \text{\AA}\)
• D. \(2.12\ \text{\AA}\)

Answer 1

The Correct Option is C

- The centripetal force required to keep the electron in a circular orbit is provided by the Coulomb force of attraction.
\[ \frac{mv^2}{r} = \frac{Ke^2}{r^2} \]
- Solving for \( v \):
\[ mv^2 = \frac{Ke^2}{r} \Rightarrow v^2 = \frac{Ke^2}{mr} \Rightarrow v = \sqrt{\frac{Ke^2}{mr}} \]

Answer 2

- The total energy \( E \) of the electron in a hydrogen atom is the sum of kinetic energy (K.E.) and potential energy (P.E.).
- Kinetic energy: \[ \text{K.E.} = \frac{1}{2}mv^2 \]
- From the centripetal force equation, we earlier derived: \[ v^2 = \frac{Ke^2}{mr} \Rightarrow \text{K.E.} = \frac{1}{2} m \cdot \frac{Ke^2}{mr} = \frac{Ke^2}{2r} \]
- Potential energy between two opposite charges: \[ \text{P.E.} = -\frac{Ke^2}{r} \]
- Total energy: \[ E = \text{K.E.} + \text{P.E.} = \frac{Ke^2}{2r} - \frac{Ke^2}{r} = -\frac{Ke^2}{2r} \]

Answer 3

- Energy of a photon emitted is given by: \[ E = \frac{hc}{\lambda} \]
- Where \( h = 6.63 \times 10^{-34} \, \text{Js} \), \( c = 3 \times 10^8 \, \text{m/s} \), and \( \lambda = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m} \)
\[ E = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{500 \times 10^{-9}} = 3.978 \times 10^{-19} \, \text{J} \]
- Convert this to electron volts: \[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}, \quad E = \frac{3.978 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.48 \, \text{eV} \]
- In Bohr's model, change in total energy \( \Delta E = -2.48 \, \text{eV} \) (as energy is released)
- Kinetic energy always equals \( -E \) (from total energy): \[ \Delta KE = -\frac{1}{2} \Delta PE = +2.48 \, \text{eV} \]

Answer 4

- In Bohr's model, the frequency of revolution \( f \) is given by: \[ f = \frac{v}{2\pi r} \]
- Using Bohr's results for velocity and radius: \[ v \propto \frac{1}{n}, \quad r \propto n^2 \Rightarrow f \propto \frac{1/n}{n^2} = \frac{1}{n^3} \]

Answer 5

- The energy of an electron in the \(n^{\text{th}}\) orbit of hydrogen is given by: \[ E_n = -13.6 \frac{1}{n^2} \text{ eV} \]
- Given energy level: \(E = -3.4\ \text{eV}\), so: \[ -3.4 = -13.6 \cdot \frac{1}{n^2} \Rightarrow \frac{1}{n^2} = \frac{3.4}{13.6} = \frac{1}{4} \Rightarrow n = 2 \]
- Radius of orbit in Bohr's model: \[ r_n = n^2 \cdot a_0, \text{ where } a_0 = 0.53\ \text{\AA} \Rightarrow r_2 = 2^2 \cdot 0.53 = 4 \cdot 0.53 = 2.12\ \text{\AA} \]
- Radius in ground state \(r_1 = 0.53\ \text{\AA}\), so change = \[ 2.12 - 0.53 = 1.59\ \text{\AA} \]