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**Einstein's** Explanation Of **Photoelectric Effect** - **Threshold Frequency** ...

Learn about **Einstein's Theory** of **Photoelectric Effect** at BYJU'S. ... did not pursue
the **matter** as he felt sure that it could be explained by the **wave theory**. ... This
implies that the **kinetic energy** of electrons increases **with light** intensity. ... Thus,
the **energy** of a **photon with** this **frequency** must be the **work function** of the metal.

For more information, see **Einstein's** Explanation Of **Photoelectric Effect** - **Threshold Frequency** ...

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

Incident light frequency \upsilon = 7.21* 10^{14} Hz

Maximum speed of the ejected electrons v = 6.0 * 10^5 m/s

The threshold frequency \upsilon_o= ?

Step 1: Get an expression for the threshold frequency

K.E = h \upsilon - h\upsilon_o

Kinetic energy K.E = \frac{1}{2}mv^2

Where, m- mass of an electron (9.1* 10^{-31} ) kg

\frac{1}{2}mv^2 = h (\upsilon - \upsilon_o)

\upsilon_o = \upsilon - \frac{mv^2}{2h}

Step 2: Substitute the known values in the above equation for threshold frequency

[math] \upsilon_o = [ 7.21 * 10^{14} - \frac{9.1* 10^{-31} * (6.0 * 10^5 )^2}{2*6.625 * 10^{-34}}] [/math]

\upsilon_o = 4.74* 10^{14} Hz

Hence, the threshold frequency for photoemission of electrons \upsilon_o = 4.74* 10^{14} Hz