Question

light of frequency $8.4 \times 10^{14}$ hz is incident on a metal surface. the stopping potential for this metal at this frequency is 2.8 v. what is the cutoff frequency of this metal?\
\
$\circ$ $3.2 \times 10^{14}$ hz\
$\circ$ $8.4 \times 10^{14}$ hz.\
$\circ$ $4.5 \times 10^{14}$ hz\
$\circ$ $7.3 \times 10^{14}$ hz\
$\circ$ $1.6 \times 10^{14}$ hz

Explanation:

Step1: Recall Photoelectric Effect Formula

The photoelectric effect equation is \( eV_s = hf - hf_0 \), where \( e \) is the electron charge, \( V_s \) is the stopping potential, \( h \) is Planck's constant, \( f \) is the incident frequency, and \( f_0 \) is the cutoff (threshold) frequency. We can rearrange this to solve for \( f_0 \): \( f_0 = f - \frac{eV_s}{h} \).

Step2: Substitute Known Values

We know \( e = 1.6 \times 10^{-19} \, \text{C} \), \( h = 6.63 \times 10^{-34} \, \text{J·s} \), \( f = 8.4 \times 10^{14} \, \text{Hz} \), and \( V_s = 2.8 \, \text{V} \). First, calculate \( \frac{eV_s}{h} \):

\[ \frac{eV_s}{h} = \frac{(1.6 \times 10^{-19} \, \text{C})(2.8 \, \text{V})}{6.63 \times 10^{-34} \, \text{J·s}} \]

\[ = \frac{4.48 \times 10^{-19} \, \text{J}}{6.63 \times 10^{-34} \, \text{J·s}} \approx 6.76 \times 10^{14} \, \text{Hz} \] Wait, no, wait—wait, let's recalculate that. Wait, \( eV_s \) is the kinetic energy, so \( hf_0 \) is the work function. Wait, maybe I made a miscalculation. Wait, \( e = 1.6e-19 \), \( V_s = 2.8 \), so \( eV_s = 1.6e-19 * 2.8 = 4.48e-19 \, \text{J} \). Then divide by \( h = 6.63e-34 \, \text{J·s} \): \( 4.48e-19 / 6.63e-34 ≈ (4.48 / 6.63) * 1e15 ≈ 0.6757 * 1e15 = 6.757e14 \, \text{Hz} \). Wait, but then \( f_0 = 8.4e14 - 6.757e14 ≈ 1.643e14 \, \text{Hz} \)? Wait, no, that can't be right. Wait, maybe I messed up the exponents. Wait, \( h = 6.63e-34 \), so \( 1/h ≈ 1.508e33 \, \text{s/J} \). Then \( eV_s = 4.48e-19 \, \text{J} \), so \( (eV_s)/h = 4.48e-19 * 1.508e33 ≈ 4.48 * 1.508e14 ≈ 6.756e14 \, \text{Hz} \). Then \( f_0 = 8.4e14 - 6.756e14 ≈ 1.644e14 \, \text{Hz} \), which is approximately \( 1.6 \times 10^{14} \, \text{Hz} \)? Wait, no, the options include 1.6e14, but let's check again. Wait, maybe my formula is wrong. Wait, the photoelectric equation is \( K_{max} = hf - \phi \), and \( K_{max} = eV_s \), so \( eV_s = hf - hf_0 \), so \( hf_0 = hf - eV_s \), so \( f_0 = f - \frac{eV_s}{h} \). Let's recalculate \( \frac{eV_s}{h} \):

\( e = 1.6e-19 \), \( V_s = 2.8 \), so \( eV_s = 4.48e-19 \, \text{J} \).

\( h = 6.626e-34 \, \text{J·s} \) (using a more precise value of Planck's constant).

So \( \frac{eV_s}{h} = 4.48e-19 / 6.626e-34 ≈ (4.48 / 6.626) * 1e15 ≈ 0.676 * 1e15 = 6.76e14 \, \text{Hz} \).

Then \( f_0 = 8.4e14 - 6.76e14 = 1.64e14 \, \text{Hz} \), which is approximately \( 1.6 \times 10^{14} \, \text{Hz} \). Wait, but let's check the options. The options are 3.2e14, 8.4e14, 4.5e14, 7.3e14, 1.6e14. So 1.6e14 is an option. Wait, but maybe I made a mistake in the calculation. Wait, let's check with \( h = 4.136e-15 \, \text{eV·s} \) (since sometimes Planck's constant is used in eV·s for easier calculation with electron volts). Let's try that. \( h = 4.136e-15 \, \text{eV·s} \). Then \( \frac{V_s}{h} \) (since \( eV_s = hf_0 \) in eV? Wait, no, \( K_{max} = eV_s = hf - hf_0 \), so \( f_0 = f - \frac{eV_s}{h} \). If we use \( h = 4.136e-15 \, \text{eV·s} \), then \( \frac{eV_s}{h} = \frac{V_s}{h/e} \), but actually, \( h = 4.136e-15 \, \text{eV·s} \), so \( \frac{eV_s}{h} = \frac{V_s}{h/e} \)? No, wait, \( e = 1.6e-19 \, \text{C} \), \( h = 4.136e-15 \, \text{eV·s} \), so \( h = 4.136e-15 \, \text{eV·s} = 4.136e-15 * 1.6e-19 \, \text{J·s} = 6.6176e-34 \, \text{J·s} \), which is close to the previous value. So using \( h = 4.136e-15 \, \text{eV·s} \), then \( \frac{V_s}{h} = \frac{2.8 \, \text{V}}{4.136e-15 \, \text{eV·s}} \) (since \( eV_s \) is in eV when \( V_s \) is in volts, because \( e*V_s \) is energy in eV). Wait, yes! If \( V_s \) is in volts, then \( eV_s \) is in eV. So \( K_{max} = eV_s = 2.8 \, \text{eV} \). Then \( hf \) is the energy of the photon in eV: \( hf = (4.136e-15 \, \text{eV·s})(8.4e14 \, \text{Hz}) = 4.136e-15 * 8.4e14 \, \text{eV} = 4.136 * 0.84 \, \text{eV} ≈ 3.474 \, \text{eV} \). Then the work function \( \phi = hf_0 = hf - K_{max} = 3.474 \, \text{eV} - 2.8 \, \text{eV} = 0.674 \, \text{eV} \). Then \( f_0 = \frac{\phi}{h} = \frac{0.674 \, \text{eV}}{4.136e-15 \, \text{eV·s}} ≈ 1.63e14 \, \text{Hz} \), which is approximately \( 1.6 \times 10^{14} \, \text{Hz} \). So that matches the earlier calculation. Therefore, the cutoff frequency is approximately \( 1.6 \times 10^{14} \, \text{Hz} \).

Answer:

1.6×10¹⁴ Hz (the option with "1.6×10¹⁴ Hz")