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Physics Class 12th Nuclei physics ncert exemplar solutions class 12th chapter two

The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 6630 $\overset{o}{A}$ is 0.42 V. If the threshold frequency is x × 10¹³ /s, where x is ________(nearest integer).

(Given, speed light = 3 × 10⁸ m/s, Planck’s constant = 6.63 × 10^-34 Js)

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Vishal Baghel | Contributor-Level 10

4 months ago

$e V_{0} + h υ_{0} = \frac{h C}{λ}$

$\Rightarrow 1.6 \times 1 0^{- 19} \times 0.42 + 6.63 \times 1 0^{- 34} υ_{0} = \frac{6.63 \times 1 0^{- 34} \times 3 \times 1 0^{8}}{66 3^{0} \times 1 0^{- 10}}$

$\Rightarrow 6.63 \times 1 0^{- 34} υ_{0} = 1 0^{- 19} (3 - 1.6 \times 0.42)$

$6.63 \times 1 0^{- 34} υ_{0} = 2.328 \times 1 0^{- 19}$

$υ_{0} = 35.11 \times 1 0^{13} = 35$

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The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 6630 $\overset{o}{A}$ is 0.42 V. If the threshold frequency is x × 10¹³ /s, where x is ________(nearest integer).

(Given, speed light = 3 × 10⁸ m/s, Planck’s constant = 6.63 × 10^-34 Js)

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The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 6630 A o is 0.42 V. If the threshold frequency is x × 1013 /s, where x is ________(nearest integer). (Given, speed light = 3 × 108 m/s, Planck’s constant = 6.63 × 10-34 Js)

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The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 6630 $\overset{o}{A}$ is 0.42 V. If the threshold frequency is x × 10¹³ /s, where x is ________(nearest integer).

(Given, speed light = 3 × 10⁸ m/s, Planck’s constant = 6.63 × 10^-34 Js)