A photon of wavelength 3310 A falls on a photo cathode and an electron of energy 3 x 10
wayelength of incident photon is charged to 5000A the energy of ejected electron is 9.72 x 10
J is ejected if the
calculate the value
of Planck constant and threshold wavelength of the photon.
Answers
The value of Planck constant and threshold wavelength of the photon is 6.62 * 10⁻³⁴J.S and 6620 * 10⁻¹⁰ m respectively.
Explanation:
=> By using the equation,
hc/λ =W₀ + kinetic energy
=> Here, A photon of wavelength 3310 Å falling on a photo cathode ejects an electron of energy 3 × 10⁻¹⁹ J.
∴ h*3*10 ⁸/3310 * 10⁻¹⁰ = W₀ + 3 * 10⁻¹⁹ ...(1)
=> one of wavelength 5000 Å ejects an electron of energy 9.72 * 10⁻²⁰ (0.972 × 10⁻¹⁹ J)
∴ h*3*10 ⁸/5000 * 10⁻¹⁰ = W₀ + 0.972 * 10⁻¹⁹ ...(2)
=> By subtracting eq (2) from (1), we get
h = 6.62 * 10⁻³⁴ J.S
=> Substituting the value of h in eq(1), we get
6.62 * 10⁻³⁴ * 3 * 10 ⁸ / 3310 * 10⁻¹⁰ = W₀ + 3 * 10⁻¹⁹
0.006 * 10⁻¹⁶ = W₀ + 3 * 10⁻¹⁹
6 * 10⁻¹⁹ = W₀ + 3 * 10⁻¹⁹
∴ W₀ = 3 * 10⁻¹⁹ J
=> Now, threshold wavelength λ₀ of the photon:
λ₀ = hc / W₀
= 3 * 10⁸ * 6.62 * 10⁻³⁴ / 3 * 10⁻¹⁹
= 6.62 * 10⁻⁷
= 6620 * 10⁻¹⁰ m.
Thus, The value of Planck constant and threshold wavelength of the photon is 6.62 * 10⁻³⁴J.S and 6620 * 10⁻¹⁰ m respectively.
Learn more:
Q:1 The value of Planck's Constant 6.63* 10^-34Js. The speed of light is3*10^17nm/s. Which value is closest to the wavelength in Nano metre of a quantum of light with frequency of 6*10^15/s?
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Q:2 If threshold wavelength for ejection of electron from metal is 330nm, then work function for photoelectric emission is?
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Answer:By using the equation,
hc/λ =W₀ + kinetic energy
=> Here, A photon of wavelength 3310 Å falling on a photo cathode ejects an electron of energy 3 × 10⁻¹⁹ J.
∴ h*3*10 ⁸/3310 * 10⁻¹⁰ = W₀ + 3 * 10⁻¹⁹ ...(1)
=> one of wavelength 5000 Å ejects an electron of energy 9.72 * 10⁻²⁰ (0.972 × 10⁻¹⁹ J)
∴ h*3*10 ⁸/5000 * 10⁻¹⁰ = W₀ + 0.972 * 10⁻¹⁹ ...(2)
=> By subtracting eq (2) from (1), we get
h = 6.62 * 10⁻³⁴ J.S
=> Substituting the value of h in eq(1), we get
6.62 * 10⁻³⁴ * 3 * 10 ⁸ / 3310 * 10⁻¹⁰ = W₀ + 3 * 10⁻¹⁹
0.006 * 10⁻¹⁶ = W₀ + 3 * 10⁻¹⁹
6 * 10⁻¹⁹ = W₀ + 3 * 10⁻¹⁹
∴ W₀ = 3 * 10⁻¹⁹ J
=> Now, threshold wavelength λ₀ of the photon:
λ₀ = hc / W₀
= 3 * 10⁸ * 6.62 * 10⁻³⁴ / 3 * 10⁻¹⁹
= 6.62 * 10⁻⁷
= 6620 * 10⁻¹⁰ m.
HENCE, The value of Planck constant and threshold wavelength of the photon is 6.62 * 10⁻³⁴J.S and 6620 * 10⁻¹⁰ m respectively.
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