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Grade 11Modern Physics

The wavelength of photons in two cases are 4000 A

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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8 Years agoGrade 11
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2 Answers

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Eshan

8 Years ago
Dear student,

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

Hence stopping potential of4000A^{\circ}photon is\dfrac{hc}{e\lambda}=3.1V


Hence stopping potential of3600A^{\circ} photon is\dfrac{hc}{e\lambda}=3.44V
Hence the required difference=0.34V
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ANJALI KUMARI

7 Years ago
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