First slide

Law of conservation of angular momentum

Question

The photoelectric threshold wavelength of silver is $3250 \times 10^{- 10} m$ . The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength $2536 \times 10^{- 10} m is$
$[Given h = 4.14 \times 10^{- 15} eVs and c = 3 \times 10^{8} m s^{- 1}]$

Moderate

A

$\approx 0.6 \times 10^{6} m s^{- 1}$
B

$\approx 61 \times 10^{3} m s^{- 1}$
C

$\approx 0.3 \times 10^{6} m s^{- 1}$
D

$\approx 6 \times 10^{5} m s^{- 1}$

Solution

$K_{\max} = h v - ϕ_{0} = h v - h v_{o} = \frac{h c}{λ} - \frac{h c}{λ_{0}} where λ_{0} = threshold wavelength or \frac{1}{2} m v^{2} = \frac{h c}{λ} - \frac{h c}{λ_{0}} \begin{array}{l} Here, h = 4.14 \times 10^{- 15} eVs, c = 3 \times 10^{8} m s^{- 1} \\ λ_{o} = 3250 \times 10^{- 10} m = 3250 \overset{0}{A} \\ λ = 2536 \times 10^{- 10} m = 2536 \overset{0}{A}, m = 9.1 \times 10^{- 31} kg \\ h c = 4.14 \times 10^{- 15} eVs \times 3 \times 10^{8} m s^{- 1} = 12420 eV \ AA \\ ∴ \frac{1}{2} m v^{2} = 12420 [\frac{1}{2536} - \frac{1}{3250}] eV = 1.076 eV \\ v^{2} = \frac{2.152 eV}{m} = \frac{2.152 \times 1.6 \times 10^{- 19}}{9.1 \times 10^{- 31}} \\ ∴ v \approx 6 \times 10^{5} m s^{- 1} = 0.6 \times 10^{6} m s^{- 1} \end{array}$

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