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 $\boxed{1 × 10^6 \ ms^{-1}}$