Question

the threshold frequency of a metal is 1.11×10^16 hz.what is the maximum kinetic energy of the photo electron produced by applying a light of 15 A on the metal ?​

Answer 1

QUESTION :-

If threshold frequency of a metal is 1.11 × 10⁻⁶ Hz Then the maximum kinetic energy of photo electrons produced by a light of wavelength 15 A⁰ on the metal is

GIVEN :-

• Threshold frequency of a metal = 1.11 × 10⁶Hz

• Wavelength of light = 15A⁰

TO FIND :-

• Maximum Kinetic energy of the photo electron

SOLUTION :-

$\large \underline{\bold {\boxed{ \bigstar{ \green{ \sf \: { {\upsilon}_{0} = \frac{c}{ {\lambda}_{0}} }}}}}}$

Where ,

• υₒ is threshold frequency
• λₒ is threshold wavelength
• c is velocity of light

We have ,

• υ₀ = 1.11 × 10⁶ s⁻¹
• c = 3 × 10⁻⁸ m/s

$\implies \sf \: 1.11 \times {10}^{6} \: \cancel{{s}^{ - 1}} = \frac{3 \times {10}^{8} \: m \cancel{ {s}^{ - 1} } }{{\lambda}_0 } \\ \\ \implies \sf \: (1.11 \times {10}^{6} )({\lambda}_0) = 3 \times {10}^{8} m \\ \\ \implies \sf \: {\lambda}_0 = \frac{3 \times {10}^{8} \: m}{1.11 \times {10}^{6} } \\ \\ \implies {\underline{ \boxed{ \blue {\sf{{\lambda}_0 = 270 \: m}}}}}$

$\large {\underline {\boxed{ \bigstar{ \green{ \sf{ \: 1Hz = s {}^{ - 1} }}}}}}$

$\large \underline{\bold {\boxed{ \bigstar{ \green{ \sf \: { 1A {}^{0} = {10}^{ - 10} m}}}}}}$

Relation between KE , λ , h , c and λ₀ is given by ,

$\large \underline{\bold {\boxed{ \bigstar{ \green{ \sf \: {KE = hc \bigg( \frac{1}{ \lambda} - \frac{1}{{ \lambda}_0} \bigg)}}}}}}$

Where,

• h is planck's constant
• c is velocity of light
• λ is wavelength
• λ₀ is threshold wavelength

We have,

• h = 6.625 × 10⁻³⁴ J.sec
• c = 3 × 10⁸ m/s
• λ = 15 × 10⁻¹⁰ m
• λ₀ = 270 m

$\implies \sf \: KE = 6.625 \times 10 {}^{ - 34} \times 3 \times 10 {}^{8} \bigg( \frac{1}{15 \times 10 {}^{ - 10} } - \frac{1}{270} \bigg) \\ \\ \implies \sf \: KE = 19.875 \times 10 {}^{ - 28} \bigg( \frac{10 {}^{ 10} }{15} - \frac{1}{270} \bigg) \\ \\ \implies \sf \: KE = 19.875 \times {10}^{ - 28} \bigg( \frac{18 \times {10}^{10} - 1 }{270} \bigg) \\ \\ \implies \sf \: KE = 19.875 \times {10}^{ - 28} \bigg( \frac{1.79 \times 10 {}^{ 11} }{270} \bigg) \\ \\ \implies \sf \: KE = \bigg( \frac{19.875 \times 1.79 \times 10 {}^{ - 17} }{270} \bigg) \\ \\ \implies \sf \: KE = \bigg( \frac{35.5 \times 10 {}^{ - 17} }{270} \bigg) \\ \\ \implies {\underline {\boxed {\blue{\sf {\: KE = 0.13 \times {10}^{ - 17}\:J}}}}}$

∴ The Kinetic energy of the photoelectrons is 0.13 × 10⁻¹⁷ J

Answer 2

Answer:

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