The threshold frequency for a certain metal is 3.3 * 10^{14} Hz. If light of frequency 8.2 * 10^{14} Hz is incident on the metal, predict the cutoff voltage for the photoelectric emission.

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CBSE Class 12th Physics Notes: Dual Nature of Radiation and ...
Jan 19, 2017 ... CBSE class 12 chapter wise notes based on chapter 11, Dual Nature of Radiation and Matter, of class 12 NCERT Physics textbook are available in this article. ... Photoelectric Effect and Wave Theory of Light. Photon. Einstein's Photoelectric Equation: Energy Quantum of Radiation. Threshold frequency.
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The difference between incident photon energy and the work function of the metal or material is the gain of kinetic energy of an electron, which can be expressed as follows
K.E = h\upsilon - \phi
Where, h - Planck's constant (6.626*10^{-34}), \phi - work function, and \upsilon - frequency
Relation between work function and threshold frequency
\phi = h \upsilon_o
Where, \upsilon_o - threshold frequency
Given that
Threshold frequency of the metal \upsilon_o = 3.3* 10^{14} Hz
Frequency of light incident on the metal \upsilon = 8.2 * 10^{14} Hz
Step 1: Set up an equation for the cutoff voltage for the photoelectric emission
K.E = h\upsilon - \phi
K.E = h\upsilon - h \upsilon_o
According to the conversion law
K.E = eV
Where, charge of electron e = 1.6* 10^{-19} and V - potential difference(voltage)
eV = h\upsilon - h \upsilon_o
V = \frac{ h(\upsilon - \upsilon_o)}{e}
Step 2: Plug in the given values in the above equation
V = \frac{h(\upsilon - \upsilon_o)}{e}
V = \frac{ 6.626 * 10^{-34} (8.2 * 10^{14} - 3.3* 10^{14})}{1.6* 10^{-19}}
V = 2.0292 volts
Therefore, the cutoff voltage for the photoelectric emission V = 2.0292 volts