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SashulF [63]
3 years ago
15

For a photoelectric tube, calculate the voltage which will be just sufficient to stop electrons emitted by the sodium photo-plat

e reaching the collector plate when light of frequency 6 x 1014Hz is incident on a sodium plate. Addition information: 1 e = 1.6 x 10-19C Plank's Constant, h = 6.64 x 10-34 Work function, Wo, of sodium = 2.2 x 10-19J
Physics
1 answer:
Talja [164]3 years ago
8 0

Answer:

1.11 V

Explanation:

Given that the Einstein photoelectric equation states that;

KE = E - Wo

E = energy of incident photon

Wo= work function of the metal

E = hf = 6.64 x 10-34 * 6 x 1014

E = 39.84 * 10^-20 J or 3.98  * 10^-19 J

KE = 3.98  * 10^-19 J - 2.2 x 10-19J

KE = 1.78 *  10^-19J

We convert this value of KE to electron volts

KE = 1.78 *  10^-19J/1.6 x 10-19C

KE = 1.11 eV

Hence; 1.11 V will be just sufficient to stop electrons emitted by the sodium photo-plate reaching the collector plate.

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m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

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m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}

v_{1}=u_{1}+u_{2}

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v_{1}=2.2i-0.80i

v_{1}=1.4i\ m/s

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m_{1}v_{1}=-m_{2}v_{2}

v_{1}=-v_{2}

Put the value into the formula

v_{1}=-1.36j\ m/s

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v_{1}=\sqrt{(1.4)^2+(1.36)^2}

v_{1}=1.95\ m/s

(b) We need to calculate the direction of the first ball after the collision

Using formula of direction

\tan\theta=\dfrac{v_{2}}{v_{1}}

\tan\theta=\dfrac{-1.36}{1.4}

\theta=\tan^{-1}\dfrac{-1.36}{1.4}

\theta=-44.16^{\circ}

Negative sign shows the direction of first ball .

Hence, (a). The speed of the first ball after the collision is 1.95 m/s.

(b). The direction of the first ball after the collision is 44.16° due south of east.

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