Question

For a certain metal, v is five times the y, and the marmum velody of coming out photo electrons is 8 x
10° Ifv = 21, the maximum velocity of photo electrons wil be
(1) 4 * 10% mis
2)610 mis
(3) 8 * 106 mie
(4) 1 2 10 ms​

Answer 1

Answer:

610 mis is the answer mate

Answer 2

The maximum velocity of photo electrons is $4\times10^{6}\ m/s$.

(1) is correct option.

Explanation:

Given that,

Velocity $v= 8\times10^{6}\ m/s$

$\nu = 5\nu_{0}$

$\nu=2\nu_{0}$

We need to calculate the maximum velocity of photo electrons

Using formula of maximum kinetic energy

$\dfrac{1}{2}mv^2=h(\nu-\nu_{0})$

Put the value into the formula

Case (I),

$\dfrac{1}{2}m(8\times10^{6})^2=h(5\nu_{0}-\nu_{0})$...(I)

Case (II),

$\dfrac{1}{2}mv^2=h(2\nu_{0}-\nu_{0})$....(II)

Divided equation (II) by (I)

$\dfrac{v^2}{(8\times10^{6})^2}=\dfrac{2\nu_{0}-\nu_{0}}{5\nu_{0}-\nu_{0}}$

$v^2=(8\times10^{6})^2\times\dfrac{1}{4}$

$v=4\times10^{6}\ m/s$

Hence, The maximum velocity of photo electrons is $4\times10^{6}\ m/s$.

Learn more :

Topic : kinetic energy

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