A metal surface is exposed to 500 nm radiation. The threshold frequency of the metal for photoelectric current is 4.3 \(\times\) 10\(^{14}\) Hz. The velocity of ejected electron is ___________ \(\times\) 10\(^{5}\) ms\(^{-1}\). (Nearest integer)
[Use: h = 6.63\(\times\)10\(^{-34}\) Js, m\(_e\) = 9.0\(\times\)10\(^{-31}\) kg]
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For photoelectric effect calculations, ensure all units are in the SI system (Joules for energy, meters for wavelength, kg for mass). A useful shortcut for photon energy in eV is \(E(eV) = \frac{1240}{\lambda(nm)}\). You can use this to calculate E and \(\phi\) in eV, find KE in eV, then convert KE to Joules (\(1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}\)) before calculating velocity.
Step 1: Understanding the Question
This question is about the photoelectric effect. We are given the wavelength of incident radiation and the threshold frequency of the metal. We need to calculate the velocity of the ejected electron. Step 2: Key Formula or Approach
The photoelectric effect is described by Einstein's equation:
\[ \text{Energy of incident photon (E)} = \text{Work function (\(\phi\))} + \text{Kinetic energy of ejected electron (KE)} \]
Where:
\(E = \frac{hc}{\lambda}\)
\(\phi = h\nu_0\) (Work function, where \(\nu_0\) is the threshold frequency)
\(KE = \frac{1}{2}m_ev^2\)
So, \(\frac{hc}{\lambda} = h\nu_0 + \frac{1}{2}m_ev^2\) Step 3: Detailed Calculation Calculate the energy of the incident photon (E):
\(\lambda = 500 \text{ nm} = 500 \times 10^{-9} \text{ m}\)
\(c = 3 \times 10^8 \text{ m/s}\)
\[ E = \frac{(6.63 \times 10^{-34} \text{ Js}) \times (3 \times 10^8 \text{ m/s})}{500 \times 10^{-9} \text{ m}} = \frac{19.89 \times 10^{-26}}{500 \times 10^{-9}} = 0.03978 \times 10^{-17} = 3.978 \times 10^{-19} \text{ J} \]
Calculate the work function (\(\phi\)):
\(\nu_0 = 4.3 \times 10^{14} \text{ Hz}\)
\[ \phi = (6.63 \times 10^{-34} \text{ Js}) \times (4.3 \times 10^{14} \text{ s}^{-1}) = 28.509 \times 10^{-20} = 2.851 \times 10^{-19} \text{ J} \]
Calculate the kinetic energy (KE) of the electron:
\[ KE = E - \phi = (3.978 - 2.851) \times 10^{-19} \text{ J} = 1.127 \times 10^{-19} \text{ J} \]
Calculate the velocity (v) of the electron:
\(KE = \frac{1}{2}m_ev^2 \Rightarrow v = \sqrt{\frac{2 \times KE}{m_e}}\)
\[ v = \sqrt{\frac{2 \times (1.127 \times 10^{-19} \text{ J})}{9.0 \times 10^{-31} \text{ kg}}} = \sqrt{\frac{2.254 \times 10^{-19}}{9.0 \times 10^{-31}}} = \sqrt{0.2504 \times 10^{12}} = \sqrt{25.04 \times 10^{10}} \]
\[ v \approx 5.0 \times 10^5 \text{ m/s} \]
Step 4: Final Answer
The velocity is 5 \(\times\) 10\(^{5}\) ms\(^{-1}\). The integer value is 5.
