An experimental study of the photoelectric effect involves a metal of work function $\phi_0$. What is the smallest wavelength of the incident photon to photoemit an electron of mass $m$ which has the same de Broglie wavelength as that of the incident photon? [Given $h$ is the Planck's constant, $c$ is the speed of light, and $\phi_0 \ll mc^2$]
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In algebraic physics problems with quadratic equations, always correlate "smallest value of variable" with the "largest value of its reciprocal" to determine whether to use the positive or negative root.
Step 1: Understanding the Question:
We are studying the photoelectric effect where a photon of wavelength $\lambda$ falls on a metal surface of work function $\phi_0$.
An electron of mass $m$ is emitted. The de Broglie wavelength of this emitted electron ($\lambda_e$) is equal to the wavelength of the incident photon ($\lambda$).
We need to find the smallest wavelength $\lambda$ of the incident photon that satisfies this condition. Step 2: Key Formula or Approach:
The energy of the incident photon is:
\[ E_{ph} = \frac{hc}{\lambda} \]
The de Broglie wavelength of the electron is $\lambda_e = \frac{h}{p}$, which means its momentum is:
\[ p = \frac{h}{\lambda_e} = \frac{h}{\lambda} \]
Since $\phi_0 \ll mc^2$, the kinetic energy of the electron can be written using non-relativistic mechanics:
\[ K = \frac{p^2}{2m} = \frac{h^2}{2m\lambda^2} \]
According to Einstein's photoelectric equation:
\[ E_{ph} = K + \phi_0 \implies \frac{hc}{\lambda} = \frac{h^2}{2m\lambda^2} + \phi_0 \]
Step 3: Detailed Explanation:
• Let us rewrite the equation in terms of $\frac{1}{\lambda}$ to form a quadratic equation:
\[ \frac{h^2}{2m} \left(\frac{1}{\lambda}\right)^2 - hc \left(\frac{1}{\lambda}\right) + \phi_0 = 0 \]
• Let $x = \frac{1}{\lambda}$:
\[ \frac{h^2}{2m} x^2 - hc x + \phi_0 = 0 \]
• Solving this quadratic equation for $x$ using the quadratic formula:
\[ x = \frac{hc \pm \sqrt{(hc)^2 - 4 \left( \frac{h^2}{2m} \right) \phi_0}}{2 \left( \frac{h^2}{2m} \right)} \]
\[ x = \frac{hc \pm \sqrt{h^2c^2 - \frac{2h^2\phi_0}{m}}}{\frac{h^2}{m}} \]
• Factoring out $h^2c^2$ from the square root:
\[ x = \frac{hc \pm hc \sqrt{1 - \frac{2\phi_0}{mc^2}}}{\frac{h^2}{m}} \]
\[ x = \frac{mc}{h} \left( 1 \pm \sqrt{1 - \frac{2\phi_0}{mc^2}} \right) \]
• For the smallest wavelength $\lambda$, we must choose the maximum possible value of $x = \frac{1}{\lambda}$. Therefore, we take the positive sign:
\[ \frac{1}{\lambda} = \frac{mc}{h} \left( 1 + \sqrt{1 - \frac{2\phi_0}{mc^2}} \right) \]
• Taking the reciprocal to find $\lambda$:
\[ \lambda = \frac{h}{mc} \left( 1 + \sqrt{1 - \frac{2\phi_0}{mc^2}} \right)^{-1} \]
Step 4: Final Answer:
The smallest wavelength of the incident photon is $\frac{h}{mc} \left( 1 + \sqrt{1 - \frac{2\phi_0}{mc^2}} \right)^{-1}$.
