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ahrayia [7]
3 years ago
5

A beam of light, with a wavelength of 4170 Å, is shined on sodium, which has a work function (binding energy) of 4.41 × 10 –19 J

. Calculate the kinetic energy and speed of the ejected electron.
Physics
1 answer:
Lyrx [107]3 years ago
7 0

Explanation:

Given that,

Wavelength of the light, \lambda=4170\ A=4170\times 10^{-10}\ m

Work function of sodium, W_o=4.41\times 10^{-19}\ J

The kinetic energy of the ejected electron in terms of work function is given by :

KE=h\dfrac{c}{\lambda}-W_o\\\\KE=6.63\times 10^{-34}\times \dfrac{3\times 10^8}{4170\times 10^{-10}}-4.41\times 10^{-19}\\\\KE=3.59\times 10^{-20}\ J

The formula of kinetic energy is given by :

KE=\dfrac{1}{2}mv^2\\\\v=\sqrt{\dfrac{2KE}{m}} \\\\v=\sqrt{\dfrac{2\times 3.59\times 10^{-20}}{9.1\times 10^{-31}}} \\\\v=2.8\times 10^5\ m/s

Hence, this is the required solution.

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uniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is station
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Answer:

The magnitude of the tangential velocity is v= 0.868 m/s

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Explanation:

From the question we are told that

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       The radius of the uniform disk is R_d = 0.200m

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Where \theta is the angular displacement given from the question as

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   \alpha is the angular acceleration which is mathematically represented as

                    \alpha = \frac{torque }{moment \ of  \ inertia}  = \frac{F_d * R_d}{I}

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Substituting values

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Substituting values

               \alph\alpha = \frac{(30.0)(0.200)}{0.8}

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Considering the equation for angular velocity

    w = \sqrt{2 \alpha  \theta}

Substituting values

     w =\sqrt{2 * (7.5) * 1.257}

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The tangential velocity of a given point on the rim is mathematically represented as

                 v = R_d w

Substituting values

                    = (0.200)(4.34)

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The radial acceleration at hat point  is mathematically represented as

            \alpha_r = \frac{v^2}{R}

                  = \frac{0.868^2}{0.200^2}

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               \alpha _t = R \alpha

Substituting values

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Substituting values

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