Given:
Step 1: Calculate photon energy (E)
Using Planck's equation:
\[ E = \frac{hc}{λ} \]
\[ E = \frac{(6.63 × 10⁻³⁴)(3 × 10⁸)}{3 × 10⁻⁷} = 6.63 × 10⁻¹⁹ J \]
Convert to eV:
\[ E = \frac{6.63 × 10⁻¹⁹}{1.6 × 10⁻¹⁹} ≈ 4.14 eV \]
Step 2: Determine kinetic energy of photoelectrons
Using Einstein's photoelectric equation:
\[ E = Φ + K.E. \]
\[ K.E. = E - Φ = 4.14 eV - 1 eV = 3.14 eV \]
Convert to joules:
\[ K.E. = 3.14 × 1.6 × 10⁻¹⁹ = 5.02 × 10⁻¹⁹ J \]
Step 3: Calculate electron velocity
Using kinetic energy formula:
\[ K.E. = \frac{1}{2}mₑv² \]
\[ v = \sqrt{\frac{2 × K.E.}{mₑ}} \]
\[ v = \sqrt{\frac{2 × 5.02 × 10⁻¹⁹}{9.11 × 10⁻³¹}} \]
\[ v = \sqrt{1.10 × 10¹²} \]
\[ v ≈ 1.05 × 10⁶ m/s \]
Step 1: Energy of Incident Photon (E)
$E = \frac{hc}{\lambda}$
$h = 6.6 \times 10^{-34} \ Js$, $c = 3 \times 10^8 \ m/s$, $\lambda = 3000 \ \AA = 3 \times 10^{-7} \ m$
$E = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{3 \times 10^{-7}} \ J = 6.6 \times 10^{-19} \ J$
Step 2: Convert Work Function to Joules ($\phi_0$)
$\phi_0 = 1 \ eV = 1.6 \times 10^{-19} \ J$
Step 3: Maximum Kinetic Energy ($KE_{max}$)
$KE_{max} = E - \phi_0 = 6.6 \times 10^{-19} \ J - 1.6 \times 10^{-19} \ J = 5 \times 10^{-19} \ J$
Step 4: Relate $KE_{max}$ to Velocity ($v_{max}$)
$KE_{max} = \frac{1}{2} m v_{max}^2$
$v_{max} = \sqrt{\frac{2 KE_{max}}{m}} = \sqrt{\frac{2 \times 5 \times 10^{-19}}{9.1 \times 10^{-31}}} \ m/s$
Step 5: Calculate $v_{max}$
$v_{max} = \sqrt{\frac{10}{9.1} \times 10^{12}} \ m/s \approx \sqrt{1.1} \times 10^6 \ m/s \approx 1 \times 10^6 \ m/s$
Step 6: Match with Options
Option (D) is $1 \times 10^6 \ ms^{-1}$.
Final Answer: The final answer is $\boxed{1 × 10^6 \ ms^{-1}}$
Einstein's Explanation of the Photoelectric Effect:
Einstein explained the photoelectric effect on the basis of Planck’s quantum theory, where light travels in the form of small bundles of energy called photons.
The energy of each photon is hν, where:
The number of photons in a beam of light determines the intensity of the incident light.When a photon strikes a metal surface, it transfers its total energy hν to a free electron in the metal.A part of this energy is used to eject the electron from the metal, and this required energy is called the work function.The remaining energy is carried by the ejected electron as its kinetic energy.