## Related questions with answers

When light of wavelength 350 nm falls on a potassium surface, electrons having a maximum kinetic energy of 1.31 eV are emitted. Find the frequency corresponding to the cutoff wavelength.

Solutions

VerifiedIn this problem, light of wavelength $\lambda = 350~\mathrm{nm}$ is incident on a potassium surface. The electrons ejected have a maximum kinetic energy of $KE_\text{max} = 1.31~\mathrm{eV}$. For this part, we calculate the cut-off frequency of the potassium's wave function.

Given values:

$\lambda=350 \ \text{nm}=350 \cdot 10^{-9} \ \text{m}$ $KE=1.31 \ \text{eV}$, $\phi=2.24 \ \text{eV}$ $\lambda=555 \ \text{nm}=5.55 \cdot 10^{-7} \ \text{m}$

In part $\textbf{c)}$ we have to find the frequency corresponding to the cutoff wavelength.

Hence, from the solution $\textit{9(b)}$ where cutoff wavelength is $\lambda_{\textrm{cutoff}}=5.55\times10^{-7}\textrm{m}$. For the wavelength corresponding frequency, it follows:

$\begin{align*} f=\frac{c}{\lambda_{\textrm{cutoff}}} \end{align*}$

Therefore, substituting initial values, we get:

$\begin{align*} f=\frac{3\times10^{8}\left(\frac{\textrm{m}}{\textrm{s}}\right)}{5.55\times10^{-7}\left(\textrm{m}\right)}=0.541\times10^{15}\frac{1}{\textrm{s}} \end{align*}$

Solution is:

$\begin{align*} \boxed{f=5.41\times10^{14}\textrm{Hz}} \end{align*}$

$f_c = \dfrac{c}{\lambda_c}$

$f_c = \dfrac{3\times 10^{8} \ m/s}{555\times 10^{-9} \ m}$

$f_c = 5.41 \times 10^{14} \ Hz$

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