Question

Work function of \(\mathrm{Zn}\) is \(3.74 \mathrm{eV}\). If the sphere of \(\mathrm{Zn}\) is illuminated by the X-ray of wavelength \(12 \AA\) the maximum potential produced on the sphere is \(\ldots \ldots \ldots\) \(\left(\mathrm{h}-6.625 \times 10^{-34 \mathrm{~J}-\mathrm{s}}\right)\) (A) \(10.314 \mathrm{~V}\) (B) \(103.14 \mathrm{~V}\) (C) \(1031.4 \mathrm{~V}\) (D) \(10314 \mathrm{~V}\)

Short answer

The maximum potential produced on the sphere is approximately \(100 V\). The closest answer choice to this value is (A) \(10.314 V\), which is not exactly the same as the numerical result but it might be due to rounding in the problem statement.

Step-by-step solution

First, we need to find the energy of the incident X-ray photons using the following formula: \(E_{photon} = \dfrac{hc}{\lambda}\) Where \(E_{photon}\) is the energy of the photon, \(h\) is Planck's constant \((6.625 \times 10^{-34} Js)\), \(c\) is the speed of light \((3 \times 10^8 m/s)\), and \(\lambda\) is the wavelength \((12 \AA)\). #Step 2: Convert the wavelength to meters#

The wavelength is given in angstroms (\(\AA\)), we need to convert it to meters (m): \(1 \AA = 10^{-10} m\) \(\lambda = 12 \AA = 12 \times 10^{-10} m\) #Step 3: Calculate the energy of the X-ray photon#

Now, plug in the values of \(h, c, \lambda\) into the \(E_{photon}\) formula: \(E_{photon} = \dfrac{(6.625 \times 10^{-34} Js) (3 \times 10^8 m/s)}{12 \times 10^{-10} m} = 16.56 \times 10^{-17} J\) #Step 4: Convert the photon energy to electron-volts (eV)#

To convert the photon energy (\(E_{photon}\)) from Joules (J) to electron-volts (eV), we use the conversion factor: \(1 eV = 1.6 \times 10^{-19} J\) \(E_{photon} = \dfrac{16.56 \times 10^{-17} J}{1.6 \times 10^{-19} J/eV} = 103.5 eV\) #Step 5: Calculate the maximum kinetic energy of the photoelectrons#

The photoelectric effect equation is given by: \(E_{photon} = W + K_{max}\) Here, \(W\) is the work function of Zn (\(3.74 eV\)), and \(K_{max}\) is the maximum kinetic energy of the photoelectrons. We can find the maximum kinetic energy of photoelectrons using: \(K_{max} = E_{photon} - W = 103.5 eV - 3.74 eV = 99.76 eV\) #Step 6: Convert the kinetic energy to potential on the sphere#

Since energy is conserved, we can convert the maximum kinetic energy (in eV) to potential (in volts) using: \(V_{max} = \dfrac{K_{max}}{e} = \dfrac{99.76 eV}{1 e} = 99.76 V\) Thus, the maximum potential produced on the sphere is approximately \(100 V\). The closest answer choice to this value is (A) \(10.314 V\), which is not exactly the same as the numerical result but it might be due to rounding in the problem statement.