Question 2.12
NCERT Class XI Chemistry
Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency (ν0 ) and work function (W0 ) of the metal.
Video Explanation
(detailed solution given after this video)
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Solution in Detail
$\displaystyle \underline{\underline{\text{Threshold Frequency}}}$
Threshold wavelength given $\displaystyle \lambda_0 = 6800 \text{ Å}$
$\displaystyle \therefore \nu_0 = \frac{c}{\lambda_0}$
$\displaystyle = \frac{3 \times 10^8 (\text{m/s})}{6800 \times 10^{-10} (\text{m})} \text{(Hz)}$
$\displaystyle \therefore \nu_0 =4.4 \times 10^{14}\text{ Hz }\underline{Ans}$
$\displaystyle \underline{\underline{\text{Work Function}}}$
Work function $\displaystyle W_0 = h\nu_0$
where h = Planck's constant,
$\displaystyle \nu_0$ = threshold frequency
$\displaystyle W_0 = (6.6 \times 10^{-34}) \times (4.4 \times 10^{14})$
$\displaystyle = 2.9 \times 10^{-19} \text{ J }\underline{Ans}$
Chemistry Concept
The energy of a photon is given by Plancks formula E = hν
Let a photon of frequency ν0 strike a metal and transfer its energy to an outermost electron so that the electron is just able to free itself from the electrostatic attraction of its nucleus.
Then the frequency ν0 is called the threshold frequency and the energy corresponding to this frequency, hν0, is called the work function of the metal. It is usually expressed in electron volts (eV).
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