What is the stopping potential?

Chemistry Level 2

If an electron has a speed of 1.0 × 1 0 4 m / s 1.0 \times 10^{4} m/s , what potential difference must be applied to stop the electron?

2.8 × 1 0 − 4 V 2.8 \times 10^{-4} V 2.3 × 1 0 − 12 J 2.3 \times 10^{-12} J 4.6 × 1 0 − 23 V 4.6 \times 10^{-23} V 5.2 × 1 0 − 1 W 5.2 \times 10^{-1} W 8.4 × 1 0 − 20 V 8.4 \times 10^{-20} V

Winod Dhamnekar by Winod DHAMNEKAR , 55, India

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1 solution

Winod Dhamnekar
Mar 26, 2020

v = 1.0 × 1 0 4 m / s , v=1.0 \times 10^4 m/s, so, v 2 = 1.0 × 1 0 8 m 2 / s 2 v^2=1.0\times 10^8 m^2/s^2 Hence K E = m e ∗ v 2 2 = 9.10938215 × 1 0 − 31 K g ∗ 1.0 × 1 0 8 m 2 / s 2 2 = 4.554691075 × 1 0 − 23 J o u l e s KE=\frac{m_e*v^2}{2}=\frac{9.10938215\times 10^{-31}Kg*1.0\times10^8m^2/s^2}{2}=4.554691075\times 10^{-23}Joules

Now K E = q e ∗ V o l t s KE=q_e*Volts so, V o l t s = K E q e = 4.554691075 × 1 0 − 23 J o u l e s 1.602176487 × 1 0 − 19 C o u l o m b = 2.8 × 1 0 − 4 V o l t s Volts=\frac{KE}{q_e}=\frac{4.554691075\times 10^{-23}Joules}{1.602176487\times 10^{-19}Coulomb}=2.8\times 10^{-4}Volts

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