Numerical problems of Planck's constant

1. 1 Numerical problems of Planck’s constant Jayam chemistry adda By, Jayam chemistry adda

2. 2 Question-1: What is the threshold energy of light radiation whose wavelength is 200 nm? If the kinetic energy of the ejected electron is 1.68 x 106J mol-1. Solution: Wavelength of photon = 200 nm According to Planck quantum law, ℎ? λ ?????? ?? ?ℎ???? = (6.626 × 10−34??) × 3 × 108?/?−1 200 × 10−9? ? = ? = 0.09939 × 10−17? Energy of one mole of photon = Energy of photon x Avogadro's number ? = 0.09939 × 10−17? × 6.023 × 1023???−1 Jayam chemistry adda ? = 0.5986 × 106?/???

3. 3 By photoelectric effect formula, we have; ½??2= ℎν − ℎν0 By substituting values, we get; ? ? 1.68 × 106 ???= 0.5986 × 106 ???− ℎν0 ? ℎν0= −1.0814 × 106 ??? Threshold energy of light radiation = 1.0814 x 106 J/mol Question-2: What is the frequency of electromagnetic radiation if the energy difference is 30 x 10-34J? Solution: Energy difference of photon = 30 x 10-34J By Planck quantum law, we have; Jayam chemistry adda

4. 4 ΔE = hν 30 x 10-34= (6.626 x 10-34) x ν ν = 4.527 s-1 Question-3: Calculate the emission rate per second of released electromagnetic radiation whose wavelength is 0.5 nm. And the power of the bulb emitting light is 20 watts. Solution: The wavelength of emitted light radiation = 0.5 nm By Planck quantum law, we have; E = hc/λ By substituting values, we get; 6.626 × 10−34? ??? 3 × 108?/??? 0.5 × 10−9? E = 39.756 x 10-17 J ? = Jayam chemistry adda

5. 5 Rate of emission of photon per second = Power of bulb / energy of photon 20 ?/? ???? = 39.756 × 10−17? Rate = 0.5030 x 1017s-1 Question-4: If the work function of a metal is 2 eV. Then what is the threshold wavelength of the light radiation? Solution: The work function of a metal is; ℎ? = 2 ?? = 2 × 1.602 × 10−19? = 3.204 × 10−19? The relationship between the wavelength and frequency is; ? =? ? ℎ? ?= 3.204 × 10−19? Jayam chemistry adda

6. 6 By substituting h and c values we get; 6.626 × 10−34???? 3 × 108????−1 3.204 × 10−19? ? = 19.878 × 10−26? 3.204 × 10−19 ? = ? = 6.20 × 10−7? ? = 620 ?? Question-5: What is the threshold frequency of the metal if the electron's binding energy is 193 J/mole? Solution: Binding energy of one mole of electrons = hν0 = 193 J/mole Binding energy of a single electron = 193/6.023 x 1023J = 32.04 x 10-23 J Jayam chemistry adda We have, E= hν0 ν0 = E/h

7. 7 ν0=32.04 x 10−23 J 6.626 × 10−34?? ν0= 4.835 x 1011s-1 Question-6: What is the mass of photon of sodium having wavelength of 550 nm? Solution: Wavelength of sodium photon = 550 nm By following de-Broglie equation, we have; ℎ λ = ?? (6.626 × 10−34??) 3 × 108?/??? 550 × 10−9? m = m=0.00401 x 10-33Kg Jayam chemistry adda m= 4.01 x 10-36Kg

8. 8 Question-7: What is the wavelength of the ball having 0.1 kg mass moving with a velocity of 20 m/sec? Solution: Mass of ball = 0.1 kg Velocity of ball = 20 m/sec According to de-Broglie law, the formula for calculating the wavelength of photon is ℎ λ = ?? 6.626 × 10−34?? 0.1 ?? × 20 ?/??? λ = λ = 3.313 x 10-34m Question-8: A gas absorbs a photon of 100 nm and emits two spectral lines. The Jayam chemistry adda wavelength of one emission line is 125 nm. What is the wavelength of the other spectral line?

9. 9 Solution: E1is the energy of the photon absorbed. E2& E3 are the energy of spectral lines emitted. Then, ?1= ?2+ ?3 According to quantum theory, ℎ? ? ?? = ℎ? ?1 =ℎ? +ℎ? ?2 ?3 1 ?1 1 ?2 +1 = ?3 1 125 ??+1 1 100 ??= ?3 0.01 ?? = 0.008 ?? +1 Jayam chemistry adda ?3

10. 10 1 ?3 = 0.01 − 0.008 ?? = 0.002 ?? ?3= 500 ?? Question-9: What is the energy of the photon corresponding to the light frequency 5 X 1015sec-1? Solution: According to the quantum theory of radiation, ? = ℎ? ? = 6.626 × 10−34???? × 5 × 1015???−1 ? = 33.13 × 10−19? Question-10:What is the angular momentum of the hydrogen electron in the fourth orbit? Jayam chemistry adda Solution:

11. 11 The Bohr’s angular momentum condition is ??? =?ℎ 2? ??? =4ℎ 2? ??? =2 × 6.626 × 10−34???? 3.14 mvr = 4.22 x 10-34J sec Jayam chemistry adda

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