The wavelength of photons in two cases are 4000 A and 3600 A respectively what is difference in stopping potential for these two? (Give solution using formula)

Question

Eshan · Answer

Dear student,

Stopping potential for an incoming photon is equal to the energy of it per unit charge.

Hence stopping potential of $4000A^{\circ}$ photon is $\dfrac{hc}{e\lambda}=3.1V$

Hence stopping potential of $3600A^{\circ}$ photon is $\dfrac{hc}{e\lambda}=3.44V$
Hence the required difference=0.34V

ANJALI KUMARI · Answer

In 1st case : stopping potential=12400/4000=3.1ev. In 2nd case : stopping potential=12400/3600=3.44ev Difference=3.44- 2.06=0.34ev

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The wavelength of photons in two cases are 4000 A and 3600 A respectively what is difference in stopping potential for these two? (Give solution using formula)

The wavelength of photons in two cases are 4000 A and 3600 A respectively what is difference in stopping potential for these two? (Give solution using formula)

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The wavelength of photons in two cases are 4000 A and 3600 A respectively what is difference in stopping potential for these two? (Give solution using formula)

The wavelength of photons in two cases are 4000 A and 3600 A respectively what is difference in stopping potential for these two? (Give solution using formula)

2 Answers

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