## Related questions with answers

Two light sources are used in a photoelectric experiment to determine the work function for a particular metal surface. When green light from a mercury lamp ($\lambda = 546.1$ nm) is used, a stopping potential of 0.376 V reduces the photocurrent to zero. What stopping potential would be observed when using the yellow light from a helium discharge tube ($\lambda = 587.5$ nm)?

Solutions

VerifiedFrom the problem $\textit{14(a)}$, we know work function of the metal is $\phi=1.90\textrm{eV}$. Thus, for the photoelectron wavelength of $\lambda=587.5\textrm{nm}$, we know that the maximum kinetic energy equation states:

$\begin{align*} K.E_{max}=\textrm{e}\left[\triangle{V}\right]=\frac{h\textrm{c}}{\lambda}-{\phi} \end{align*}$

Hence, we calculate maximum kinetic energy of the ejected electrons as:

$\begin{align*} K.E_{max}=\frac{6.63\times10^{-34}\textrm{J}\cdot{s}\left(3\times10^{8}\frac{\textrm{m}}{\textrm{s}}\right)}{587.5\times10^{-9}\textrm{m}}\left[\frac{1\textrm{eV}}{1.6\times10^{-19}\textrm{J}}\right]-1.90\textrm{eV} \end{align*}$

Solution is:

$\begin{align*} \boxed{K.E_{max}=0.216\textrm{eV}} \end{align*}$

$e \ \Delta V = \dfrac{h \ c}{\lambda} - \phi$

$e \ \Delta V = (\dfrac{(6.63 \times 10^{-34})*(3 \times 10^{8})}{587.5 \times 10^{-9} \ m}) \ (\dfrac{1 \ eV}{1.60 \times 10^{-19} \ J}) - 1.90 \ eV$

$e \ \Delta V = 0.216 \ eV$

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