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CBSE Class 12 Physics 2025 All Sets Solved Paper

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Question : 9 of 16
Marks: +1, -0
In a photoelectric experiment with a material of work function 2.1 eV, the stopping potential is found to be 2.5 V. The maximum kinetic energy of ejected photoelectrons is
Solution:  
Using the photoelectric equation
KEmax=e Vs\mathrm{KE}_{max} = e\,V_s
Where
• Vs = 2.3 V (Stopping potential)
• e=1.6×10−19 Ce = 1.6 \times 10^{-19}\,\mathrm{C} (Charge of electron)
KEmax=2.5×1=2.5 eV\mathrm{KE}_{max} = 2.5 \times 1 = 2.5\,\mathrm{eV}
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