The work function of the nickel is 5 eV. When a light of wavelength 2000 mathringA falls on it, it emits photoelectrons in the circuit. Then the potential difference necessary to stop the fastest electrons emitted is (given . h=6.67 × 10-34 J - s )

Question

The work function of the nickel is 5 eV. When a light of wavelength 2000 mathringA falls on it, it emits photoelectrons in the circuit. Then the potential difference necessary to stop the fastest electrons emitted is (given . h=6.67 × 10-34 J - s ) (A) 1.0 V (B) 1.75 V (C) 1.25 V (D) 0.75 V

Tardigrade Admin · Accepted Answer

Stopping potential is given by e V0=(h c/λ)-φ ∴ e V0=(6.67 × 10-34 × 3 × 108/2 × 10-7) -5 × 1.6 × 10-19 =10.005 × 10-19-8 × 10-19 =2.005 × 10-19 or V0 =(2.005 × 10-19/1.6 × 10-19) volt =1.25 V

Q. The work function of the nickel is 5eV. When a light of wavelength 2000A˚ falls on it, it emits photoelectrons in the circuit. Then the potential difference necessary to stop the fastest electrons emitted is (given h=6.67×10−34J−s)

1.0 V

1.75 V

1.25 V

0.75 V

Stopping potential is given by eV0​=λhc​−ϕ ∴eV0​=2×10−76.67×10−34×3×108​ −5×1.6×10−19 =10.005×10−19−8×10−19 =2.005×10−19 or V0​=1.6×10−192.005×10−19​ volt =1.25V

Q. The work function of the nickel is $5 e V$ . When a light of wavelength $2000 \overset{˚}{A}$ falls on it, it emits photoelectrons in the circuit. Then the potential difference necessary to stop the fastest electrons emitted is (given $h = 6.67 \times 1 0^{- 34} J - s)$