## 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 cutoff wavelength.

Solutions

VerifiedGiven values:

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

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

From the problem $\textit{9(a)}$ we know that the potassium work function yields $\phi=2.24\textrm{eV}$.

Therefore we can calculate cutoff wavelength $\lambda_{\textrm{cutoff}}$ as:

$\begin{align*} \lambda_{\textrm{cutoff}}=\frac{h\textrm{c}}{\phi}=\frac{6.63\times10^{-34}\left(\textrm{J}\cdot{\textrm{s}}\right)3\times10^{8}\left(\frac{\textrm{m}}{\textrm{s}}\right)}{2.24\left(\textrm{eV}\right)} \end{align*}$

It follows:

$\begin{align*} \lambda_{\textrm{cutoff}}=8.88\times10^{-26}\textrm{J}\left[\frac{1\textrm{eV}}{1.6\times10^{-19}\textrm{J}}\right] \end{align*}$

Solution is:

$\begin{align*} \boxed{\lambda_{\textrm{cutoff}}=5.55\times10^{-7}\textrm{m}} \end{align*}$

$\lambda_c = \dfrac{h \ c}{\phi}$

$\lambda_c = (\dfrac{(6.63e-34)*(3e8)}{2.24 \ eV}) (\dfrac{1 \ eV}{1.60e-19 \ J})$

$\lambda_c = 5.55 \times 10^{-7} \ eV$

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