Question

light of frequency $9.6 \times 10^{14}$ hz is incident on a metal surface. the stopping potential for this metal at this frequency is 2.2 v. how much work is required for an electron to escape the surface of this metal?

\\( \circ \\) $3.2 \times 10^{-19}$ j
\\( \circ \\) $2.2 \times 10^{-19}$ j
\\( \circ \\) $1.6 \times 10^{-19}$ j
\\( \circ \\) $3.5 \times 10^{-19}$ j
\\( \circ \\) $2.8 \times 10^{-19}$ j

Explanation:

Step1: Recall the photoelectric effect equation

The photoelectric effect equation is \( E = hf = W + eV_s \), where \( E \) is the energy of the incident photon, \( h \) is Planck's constant, \( f \) is the frequency of the light, \( W \) is the work function (work required to escape), \( e \) is the elementary charge (\( e = 1.6\times 10^{-19}\, \text{C} \)), and \( V_s \) is the stopping potential. We can rearrange this to find \( W = hf - eV_s \), but we can also think about the maximum kinetic energy of the electron \( K_{max}=eV_s \), and from the photoelectric effect, the work function \( W = hf - K_{max} \). However, another way: we know that the work function \( W \) can also be related to the stopping potential. Wait, actually, the work function is the minimum energy required to eject an electron, and the maximum kinetic energy of the ejected electron is \( K_{max}=eV_s \). From \( hf = W + K_{max} \), so \( W = hf - K_{max} \). But first, let's calculate \( hf \). Planck's constant \( h = 6.63\times 10^{-34}\, \text{J·s} \), frequency \( f = 9.6\times 10^{14}\, \text{Hz} \). So \( hf=(6.63\times 10^{-34}\, \text{J·s})\times(9.6\times 10^{14}\, \text{Hz}) \approx 6.36\times 10^{-19}\, \text{J} \). Then \( K_{max}=eV_s=(1.6\times 10^{-19}\, \text{C})\times(2.2\, \text{V}) = 3.52\times 10^{-19}\, \text{J} \). Wait, no, that can't be. Wait, maybe I made a mistake. Wait, actually, the work function \( W \) is equal to \( hf - eV_s \)? No, wait, the correct equation is \( hf = W + e V_s \), so \( W = hf - e V_s \). Wait, but let's check the values. Wait, maybe the question is simpler. Wait, the work function is the minimum energy to escape, and the stopping potential is related to the maximum kinetic energy. Wait, no, maybe I confused the formula. Wait, actually, the maximum kinetic energy of the electron is \( K_{max}=eV_s \), and from the photoelectric effect, \( hf = W + K_{max} \), so \( W = hf - K_{max} \). But let's compute \( hf \): \( h = 6.626\times 10^{-34}\, \text{J·s} \), \( f = 9.6\times 10^{14}\, \text{Hz} \), so \( hf = 6.626\times 10^{-34} \times 9.6\times 10^{14} \approx 6.36\times 10^{-19}\, \text{J} \). Then \( K_{max}=eV_s = 1.6\times 10^{-19}\, \text{C} \times 2.2\, \text{V} = 3.52\times 10^{-19}\, \text{J} \). Then \( W = 6.36\times 10^{-19} - 3.52\times 10^{-19} \approx 2.84\times 10^{-19}\, \text{J} \), which is approximately \( 2.8\times 10^{-19}\, \text{J} \). Wait, but let's check the options. The options are \( 3.2\times 10^{-19}\, \text{J} \), \( 2.2\times 10^{-19}\, \text{J} \), \( 1.6\times 10^{-19}\, \text{J} \), \( 3.5\times 10^{-19}\, \text{J} \), \( 2.8\times 10^{-19}\, \text{J} \). So the answer should be \( 2.8\times 10^{-19}\, \text{J} \). Wait, but let's verify again. Wait, maybe I messed up the formula. Wait, the work function is the energy required to escape, so when the photon energy is \( hf \), the electron gains that energy, uses \( W \) to escape, and the rest is kinetic energy \( K_{max}=eV_s \). So \( W = hf - e V_s \). Let's compute \( hf \): \( 6.626e-34 * 9.6e14 = 6.626*9.6e-20 = 63.6096e-20 = 6.36096e-19 J \). Then \( e V_s = 1.6e-19 C * 2.2 V = 3.52e-19 J \). Then \( W = 6.36096e-19 - 3.52e-19 = 2.84096e-19 J \approx 2.8e-19 J \), which matches the last option.

Step2: Calculate \( hf \)

\( h = 6.626\times 10^{-34}\, \text{J·s} \), \( f = 9.6\times 10^{14}\, \text{Hz} \)
\( hf = 6.626\times 10^{-34} \times 9.6\times 10^{14} \approx 6.36\times 10^{-19}\, \text{J} \)

Step3: Calculate \( e V_s \)

\( e = 1.6\times 10^{-19}\, \text{C} \), \( V_s = 2.2\, \text{V} \)
\( e V_s = 1.6\times 10^{-19} \times 2.2 = 3.52\times 10^{-19}\, \text{J} \)

Step4: Calculate work function \( W \)

\( W = hf - e V_s \)
\( W = 6.36\times 10^{-19} - 3.52\times 10^{-19} \approx 2.8\times 10^{-19}\, \text{J} \)

Answer:

\( 2.8\times 10^{-19}\, \text{J} \) (the option with \( 2.8\times 10^{-19}\, \text{J} \))