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

Learn about **Einstein's Theory** of **Photoelectric Effect** at BYJU'S. ... According to
the **wave theory**, **energy** is uniformly distributed across the ... However, electron
emission was spontaneous no **matter** how small the intensity of **light**. ... Thus,
**Einstein** explained the **Photoelectric effect** by using the particle **nature** of **light**.

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

Given that

Work function of caesium = 2.14 eV

Incident radiation (energy) E = hf

Where, h - Planck constant = 6.63* 10^{-34} J and f - threshold frequency of waves.

a) the threshold frequency for caesium

Threshold frequency and work function relation

Incident radiation must be equivalent to the work function

W = E

W = hf

Threshold frequency f = \frac{W}{h}

f = \frac{2.14 eV}{6.63 * 10^{-34}}

f = 5.2* 20^{14} Hz

Hence, no photoelectrons are ejected for frequencies below this threshold frequency.

b) the wavelength of the incident light if the photocurrent is brought to zero by a stopping potential of 0.60 V

Potential eV = 0.60 V

Step 1: Sep up an equation for wavelength of the incident light

Photoelectric equation

eV = hf - W

Where, eV = \text{ potential difference } , h - Planck constant = 6.63* 10^{-34} J , f - frequency of waves, and f - work function.

eV = \frac{hc}{\lambda} - W \because f=\frac{\left(c\right)\text{ speed of light }}{\left(\lambda\right)\text{Wave length }}

Wavelength, \lambda = \frac{hc}{eV + W}

Step 2: Plug in the known values in the above equation

\lambda = \frac{6.63 * 10^{-34} * 3*10^{8}}{0.6 eV + 2.14 eV}

\lambda = \frac{19.89 10^{-26}}{2.74}

\lambda = 454 nm

Hence, wavelength of the incident light \lambda = 454 nm