Question

Energy of photon having wavelength \(\lambda\) is \(2 \mathrm{eV}\). This photon when incident on metal. Maximum velocity of emitted is \(\mathrm{V}\). If \(\lambda\) is decreased \(25 \%\) and maximum velocity is made double, work function of metal is \(\ldots \ldots \ldots . . \mathrm{V}\) (A) \(1.2\) (B) \(1.5\) (C) \(1.6\) (D) \(1.8\)

Short answer

The work function of the metal is \(1.6\,eV\).

Step-by-step solution

Write down the Photoelectric effect equation

The photoelectric effect equation describes how the energy of the incident photon with a known wavelength is related to the work function (ϕ) of the metal and the maximum kinetic energy of the emitted electrons. The equation is: \(E = h\nu = h\cfrac{c}{\lambda}\) In the case of the photoelectric effect, we have: \(h\cfrac{c}{\lambda} = \phi + K_{max}\) where \(h\) is Planck's constant, \(c\) is the speed of light, \(\lambda\) is the wavelength of the incident photon, \(\phi\) is the work function of the metal, and \(K_{max}\) is the maximum kinetic energy of the emitted electrons.

Establish initial conditions

Initially given: Energy of photon: \(E = 2eV\) Wavelength of photon: \(\lambda\) Maximum emitted electron velocity: \(V\) We can write the initial equation as: \(2 = \cfrac{hc}{\lambda} - \phi\) Since the maximum kinetic energy of the emitted electrons is given by: \(K_{max} = \cfrac{1}{2}mv^2\) We have: \(2 = \cfrac{hc}{\lambda} - \cfrac{1}{2}mV^2\)

Account for decreased wavelength and increased velocity

When the wavelength is decreased by 25% (\(0.75\lambda\)) and maximum velocity is doubled (\(2V\)), we have: Energy of photon: \(E = 2eV\) Wavelength of photon: \(0.75\lambda\) Maximum emitted electron velocity: \(2V\) So, the new equation becomes: \(2 = \cfrac{hc}{0.75\lambda} - \cfrac{1}{2}m(2V)^2\) Simplify the equation: \(2 = \cfrac{4hc}{3\lambda} - 2mV^2\)

Solve for the work function

Now, we have two equations: 1) \(2 = \cfrac{hc}{\lambda} - \cfrac{1}{2}mV^2\) 2) \(2 = \cfrac{4hc}{3\lambda} - 2mV^2\) Now, we can subtract equation (1) from equation (2) to eliminate \(K_{max}\): \(0 = \cfrac{4hc}{3\lambda} - \cfrac{hc}{\lambda} - \cfrac{3}{2}mV^2\) Simplify and solve for the work function: \(\cfrac{1}{2}mV^2 = \cfrac{hc}{\lambda}\left(\cfrac{1}{3}\right)\) Since we know the photon's energy is 2eV, we can replace it in the equation, and then solve for the work function ϕ: \(2 = \cfrac{hc}{\lambda} = \cfrac{3}{2}mV^2 + \cfrac{hc}{\lambda}\left(\cfrac{1}{3}\right)\) \(2 = 1 + \cfrac{hc}{\lambda}\left(\cfrac{1}{3}\right)\) Now, solve for ϕ: \(\phi = \cfrac{hc}{\lambda}\left(\cfrac{1}{3}\right)\) Using the known constants h (Planck's constant) and c (speed of light), we find the value of \(hc \approx 1240 \, eV \cdot nm\). Substituting this expression into the equation above, we get: \(\phi = \cfrac{1240}{3\lambda}\) The options given are in eV, so we need to convert this expression into eV by dividing by 2 (since the initial energy is 2 eV): \(\phi = \cfrac{1240}{6\lambda}\) Now check which option matches the given expressions: (A) \(1.2 = \cfrac{1240}{6\lambda} \Rightarrow \lambda = 172\, nm\) (B) \(1.5 = \cfrac{1240}{6\lambda} \Rightarrow \lambda = 137\, nm\) (C) \(1.6 = \cfrac{1240}{6\lambda} \Rightarrow \lambda = 129\, nm\) (D) \(1.8 = \cfrac{1240}{6\lambda} \Rightarrow \lambda = 115\, nm\) Comparing these options with the initial equation we derived earlier \(\phi = \cfrac{hc}{\lambda} - 2\) and calculating the wavelengths for each option, we find that option (C) gives us the most accurate result fulfilling the initial conditions. So the work function of the metal is \(1.6\,eV\).