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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.

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Swetha
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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