Question

Work function of tungsten and sodium are \(4.5 \mathrm{eV}\) and $2.3 \mathrm{eV}\( respectively. If threshold wavelength for sodium is \)5460 \mathrm{~A}$, threshold frequency for tungsten will be ............ (A) \(10^{15} \mathrm{~Hz}\) (B) \(1.1 \times 10^{15} \mathrm{~Hz}\) (C) \(1.2 \times 10^{15} \mathrm{~Hz}\) (D) \(1.4 \times 10^{15} \mathrm{~Hz}\)

Short answer

The threshold frequency for tungsten is approximately \(1.1 \times 10^{15} \mathrm{~Hz}\).

Step-by-step solution

Convert work function values to Joules

The work function values for tungsten and sodium are given in eV. To convert these values to Joules (J), we use the electron volt to Joule conversion factor: \(1 \mathrm{eV} = 1.6 \times 10^{-19}\ \mathrm{J}\). Thus, we have: Work function of tungsten: \[4.5 \mathrm{eV} \times 1.6 \times 10^{-19} \mathrm{\frac{J}{eV}} = 7.2 \times 10^{-19} \mathrm{J}\] Work function of sodium: \[2.3 \mathrm{eV} \times 1.6 \times 10^{-19} \mathrm{\frac{J}{eV}} = 3.68 \times 10^{-19} \mathrm{J}\]

Convert threshold wavelength to meters

The threshold wavelength for sodium is given in Angstroms (Å). To convert it to meters (m), we use the angstrom to meter conversion factor: \(1 \mathrm{Å} = 10^{-10} \mathrm{m}\). Thus, we have: Threshold wavelength for sodium: \[5460 \mathrm{~Å} \times 10^{-10} \mathrm{\frac{m}{Å}} = 5.46 \times 10^{-7}\mathrm{m}\]

Find threshold frequency for sodium

Using the threshold wavelength of sodium, we can find the threshold frequency using the speed of light (c) and the relationship: \[c = \lambda \nu\] Where \(\lambda\) is the threshold wavelength, and \(\nu\) is the threshold frequency. Therefore, we have: Threshold frequency for sodium: \[\nu_{Na} = \frac{c}{\lambda} = \frac{3 \times 10^8 \mathrm{ \frac{m}{s}}}{5.46 \times 10^{-7}\mathrm{m}} = 5.49 \times 10^{14} \mathrm{~Hz}\]

Use Planck's constant to find threshold frequency for tungsten

Using Planck's constant (h) and the work functions, we can relate the threshold frequencies for tungsten and sodium: \[h(\nu_{W} - \nu_{Na}) = \phi_{W} - \phi_{Na}\] Where \(\nu_{W}\) is the threshold frequency for tungsten, \(\nu_{Na}\) is the threshold frequency for sodium, \(\phi_{W}\) is the work function of tungsten, and \(\phi_{Na}\) is the work function of sodium. Using the calculated values, we have: \[(6.626 \times 10^{-34} \mathrm{Js})(\nu_{W} - 5.49 \times 10^{14} \mathrm{~Hz}) = (7.2 - 3.68) \times 10^{-19} \mathrm{J}\]

Solve for threshold frequency for tungsten

Now, we can solve for the threshold frequency of tungsten: \[\nu_{W} = \frac{(7.2 - 3.68) \times 10^{-19} \mathrm{J} + 6.626 \times 10^{-34} \mathrm{Js} (5.49 \times 10^{14} \mathrm{Hz})}{6.626 \times 10^{-34} \mathrm{Js}} = 1.09 \times 10^{15}\ \mathrm{Hz}\] The threshold frequency for tungsten is approximately \(1.1 \times 10^{15} \mathrm{~Hz}\), which corresponds to option (B).