In a photoelectric experiment the stopping potentials measured are 0.49 V for light of frequency 6.0x10^14 Hz and 1.73 V for 9.0x10^14 Hz. The threshold frequency of the metal, obtained from these two readings, is closest to

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In a photoelectric experiment the stopping potentials measured are $0.49 V$ for light of frequency $6.0 \times 1 0^{14} Hz$ and $1.73 V$ for $9.0 \times 1 0^{14} Hz$ . The threshold frequency of the metal, obtained from these two readings, is closest to

A $4.8 \times 1 0^{14} Hz$
B $6.0 \times 1 0^{14} Hz$
C $5.4 \times 1 0^{14} Hz$
D $7.2 \times 1 0^{14} Hz$

Answer: A

V0 = (h/e)(nu - nu0), so the slope of V0 against nu is (1.73 - 0.49)/(3.0e14) = 4.167e-15 V s (this is h/e). At nu = 6.0e14, 0.49 = slope (6.0e14 - nu0), so nu0 = 6.0e14 - 0.49/4.167e-15 = 4.848e+14 Hz.

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In a photoelectric experiment the stopping potentials measured are 0.49 V for light of frequency 6.0×1014 Hz and 1.73 V for 9.0×1014 Hz. The threshold frequency of the metal, obtained from these two readings, is closest to

Answer: A

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In a photoelectric experiment the stopping potentials measured are $0.49 V$ for light of frequency $6.0 \times 1 0^{14} Hz$ and $1.73 V$ for $9.0 \times 1 0^{14} Hz$ . The threshold frequency of the metal, obtained from these two readings, is closest to