Question

the work function for a certain sample is 2.3 ev. the stopping potential for electrons ejected from the sample by 7.0 × 10¹⁴ hz electromagnetic radiation is

○ 5.2 v
○ 0 v
○ 2.3 v
○ 0.59 v
○ 2.9 v

Explanation:

Step1: Recall the photoelectric effect formula

The photoelectric effect equation is \( E = hf = \phi + eV_s \), where \( E \) is the energy of the incident photon, \( h \) is Planck's constant (\( h = 4.136\times 10^{-15}\, \text{eV}\cdot\text{s} \)), \( f \) is the frequency of the radiation, \( \phi \) is the work function, \( e \) is the electron charge, and \( V_s \) is the stopping potential. We can rearrange the formula to solve for \( V_s \): \( V_s=\frac{hf - \phi}{e} \). Since \( e \) is in the numerator and denominator, we can write \( V_s=\frac{hf}{e}-\frac{\phi}{e} \), and \( \frac{hf}{e} \) is the photon energy in eV, so \( V_s = E_{\text{photon}}-\phi \) (in eV, and then the unit of \( V_s \) will be volts as the work function is in eV and the photon energy calculation will give eV, and when we subtract and divide by \( e \) (but since \( \phi \) is in eV, \( \frac{\phi}{e} \) is \( \phi \) in volts? Wait, no. Wait, the work function \( \phi \) is given in eV, which is \( \phi = e\phi' \) where \( \phi' \) is in volts. Wait, actually, 1 eV is the energy gained by an electron moving through 1 volt, so \( E = qV \), so if \( E \) is in eV and \( q = e \), then \( V=\frac{E}{e} \) in volts. So let's correct:

The energy of the photon is \( E = hf \). Let's calculate \( E \) in eV. \( h = 4.136\times 10^{-15}\, \text{eV}\cdot\text{s} \), \( f = 7.0\times 10^{14}\, \text{Hz} \).

Step2: Calculate the photon energy

\( E = hf=(4.136\times 10^{-15}\, \text{eV}\cdot\text{s})\times(7.0\times 10^{14}\, \text{Hz}) \)
\( E = 4.136\times7.0\times 10^{-15 + 14}\, \text{eV} \)
\( E = 28.952\times 10^{-1}\, \text{eV}=2.8952\, \text{eV}\approx 2.9\, \text{eV} \)

Step3: Calculate the stopping potential

Using \( V_s=\frac{E - \phi}{e} \), but since \( E \) and \( \phi \) are in eV, and \( e \) is the electron charge, when we divide by \( e \), the eV units will convert to volts. So \( V_s = E-\phi \) (in eV, and the result is in volts because \( 1\, \text{eV}=e\times1\, \text{V} \), so \( \frac{\text{eV}}{e}=1\, \text{V} \)).

Given \( \phi = 2.3\, \text{eV} \), \( E\approx 2.9\, \text{eV} \)

\( V_s=2.9 - 2.3 = 0.6\, \text{V} \) (approx 0.59 V due to more precise calculation of \( E \))

Wait, let's recalculate \( E \) more precisely:

\( h = 6.626\times 10^{-34}\, \text{J}\cdot\text{s} \), to convert to eV·s, we know that \( 1\, \text{eV}=1.602\times 10^{-19}\, \text{J} \), so \( h=\frac{6.626\times 10^{-34}}{1.602\times 10^{-19}}\, \text{eV}\cdot\text{s}\approx 4.136\times 10^{-15}\, \text{eV}\cdot\text{s} \) (which matches the earlier value).

\( f = 7.0\times 10^{14}\, \text{Hz} \)

\( E = hf = 4.136\times 10^{-15}\times7.0\times 10^{14}=4.136\times7.0\times 10^{-1}=2.8952\, \text{eV}\approx 2.9\, \text{eV} \)

Then \( V_s=\frac{E - \phi}{e}=\frac{(2.8952 - 2.3)\, \text{eV}}{e}=\frac{0.5952\, \text{eV}}{e}=0.5952\, \text{V}\approx 0.59\, \text{V} \)

Answer:

0.59V