The problem is regarding the photoelectric effect, where electrons are ejected from a metal surface when illuminated by light of a certain wavelength. The key formula that governs this problem is the photoelectric equation given by Einstein:
\(E = \frac{1}{2}mv^2 + \phi\)
Where:
The energy of a photon is given by:
\(E = \frac{hc}{\lambda}\)
Where:
Initially, the wavelength is \(\lambda\), and the speed of the fastest electron is \(v\). By applying the photoelectric equation:
\(\frac{hc}{\lambda} = \frac{1}{2}mv^2 + \phi\)
When the wavelength changes to \(\frac{3}{4}\lambda\), the energy of the photon increases:
\(\frac{hc}{\frac{3}{4}\lambda} = \frac{hc}{\lambda} \times \frac{4}{3}\)
The new equation for kinetic energy becomes:
\(\frac{hc}{\lambda} \times \frac{4}{3} = \frac{1}{2}m(v')^2 + \phi\)
By subtracting the two equations, we eliminate the work function:
\(\frac{hc}{\lambda} \left(\frac{4}{3} - 1\right) = \frac{1}{2}m\left((v')^2 - v^2\right)\)
Simplifying gives:
\(\frac{hc}{\lambda} \times \frac{1}{3} = \frac{1}{2}m\left((v')^2 - v^2\right)\)
Solving for \(v'\), the speed of the fastest electron with the new wavelength:
\((v')^2 = v^2 + \frac{2}{3}\frac{hc}{\lambda m}\)
And, knowing \(\frac{hc}{\lambda} = \frac{1}{2}mv^2 + \phi\), simplifies to:
\((v')^2 = v^2 + \frac{2}{3}\times\left[\frac{1}{2}v^2\right]\)
Simplifying further, we have:
\((v')^2 = v^2 \left(1 + \frac{1}{3}\right) = v^2 \times \frac{4}{3}\)
Finally,
\(v' = v \times \sqrt{\frac{4}{3}}\)
So, the speed of the fastest emitted electron is:
\(v' < v \times \left(\frac{4}{3}\right)^{1/2}\)
Correct Answer: less than \(v\left(\frac{4}{3}\right)^{1/2}\)