Question

the binding energy of electrons to a chromium metal surface is 7.21×10^(-19) j. what is the longest wavelength of light (in nm) that will eject electrons from chromium metal? etextbook and media what frequency is required to give electrons with kinetic energy of 7.61e - 19 j? what wavelength (in nm) is required to give electrons with kinetic energy of 7.61e - 19 j?

Explanation:

Step1: Recall the photoelectric - effect equation

The photoelectric - effect equation is $E = h
u=h\frac{c}{\lambda}$, where $E$ is the energy of the photon, $h = 6.626\times10^{-34}\ J\cdot s$ is Planck's constant, $
u$ is the frequency of the light, $c = 3.0\times10^{8}\ m/s$ is the speed of light, and $\lambda$ is the wavelength of the light.

First sub - question: Longest wavelength to eject electrons

When the electron is just ejected, the kinetic energy of the ejected electron $K = 0$. The energy of the photon is equal to the binding energy $E_0$ of the electron.

Step1: Set up the energy - wavelength relation

$E_0=h\frac{c}{\lambda_{max}}$, where $E_0 = 7.21\times 10^{-19}\ J$.
We can solve for $\lambda_{max}$: $\lambda_{max}=\frac{hc}{E_0}$.

Step2: Substitute the values

$\lambda_{max}=\frac{6.626\times 10^{-34}\ J\cdot s\times3.0\times 10^{8}\ m/s}{7.21\times 10^{-19}\ J}$
$=\frac{19.878\times 10^{-26}\ J\cdot m}{7.21\times 10^{-19}\ J}=2.76\times 10^{-7}\ m$.
Convert to nanometers: $\lambda_{max}=276\ nm$.

Second sub - question: Frequency for a given kinetic energy

The photoelectric - effect equation with non - zero kinetic energy is $E = h
u=E_0 + K$.
We know $E_0 = 7.21\times 10^{-19}\ J$, $K = 7.61\times 10^{-19}\ J$, and $h = 6.626\times 10^{-34}\ J\cdot s$.

Step1: Calculate the total energy of the photon

$E=E_0 + K=(7.21\times 10^{-19}+7.61\times 10^{-19})\ J = 14.82\times 10^{-19}\ J$.

Step2: Solve for the frequency

Since $E = h
u$, then $
u=\frac{E}{h}=\frac{14.82\times 10^{-19}\ J}{6.626\times 10^{-34}\ J\cdot s}=2.24\times 10^{15}\ Hz$.

Third sub - question: Wavelength for a given kinetic energy

First, find the energy of the photon as in the second sub - question: $E=E_0 + K = 14.82\times 10^{-19}\ J$.
Since $E = h\frac{c}{\lambda}$, we can solve for $\lambda$.

Step1: Rearrange the formula for wavelength

$\lambda=\frac{hc}{E}$.

Step2: Substitute the values

$\lambda=\frac{6.626\times 10^{-34}\ J\cdot s\times3.0\times 10^{8}\ m/s}{14.82\times 10^{-19}\ J}$
$=\frac{19.878\times 10^{-26}\ J\cdot m}{14.82\times 10^{-19}\ J}=1.34\times 10^{-7}\ m$.
Convert to nanometers: $\lambda = 134\ nm$.

Answer:

1. $276$ nm
2. $2.24\times 10^{15}$ Hz
3. $134$ nm