D. Correct For an alpha particle, charge is double and mass is 4 times that of a proton. Hence this answer is correct.

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Nathan Harmon
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1 OAT Physics - Problem Drill 23: Atomic Physics Question No. 1 of The specific charge of a proton is 9.6 X 10 7 C/Kg. An alpha particle consists of two protons and two neutrons, then the specific charge of an alpha particle is. Question #01 A. 2.4 x 10 7 C/Kg B. 9.6 x 10 7 C/Kg C. 1.2 x 10 7 C/Kg D. 4.8 x 10 7 C/Kg E. 1.0 x 10 7 C/Kg A. Incorrect! For an alpha particle, charge is double and mass is 4 times that of a proton and not 8 times a proton. B. Incorrect! For an alpha particle, charge is double and mass is 4 times that of a proton and not twice that of a proton. C. Incorrect! For an alpha particle, charge is double and mass is 4 times that of a proton and not 1/16 th of a proton. D. Correct For an alpha particle, charge is double and mass is 4 times that of a proton. Hence this answer is correct. E. Incorrect Check your calculation. For an alpha particle, charge is double and mass is 4 times that of a proton. The specific charge of a proton is (e/m) p = 9.6 X 10 7 C/Kg For an alpha particle, charge is double and mass is 4 times that of a proton Hence, specific charge of an alpha particle is (e/m) p is ½ that of the proton itself. 9.6 X 10 7 C/Kg(1/2)= 4.8 X 10 7 C / Kg (D) is the correct answer choice.

2 Question No. 2 of The voltage across the electrodes of cathode ray tube is 1000 volt. Calculate the speed of the electrons. Given mass of an electron = kg and charge of an electron is C. Question #02 A x 10 7 m/s B x 10-7 m/s C x 10-3 m/s D x 10-3 m/s E. 1 m/s A. Correct! Potential difference between anode and cathode V = 1000 Volt. m = kg and q= c q / m = 1v 2 /2V v = (2qV/m) B. Incorrect Equate the kinetic energy and electrical energy. C. Incorrect You ll need to recall and use the definition of kinetic energy, KE=mv 2 /2 D. Incorrect! You ll need to recall and use the definition of voltage, V=energy/charge. D. Incorrect Equate the kinetic energy and electrical energy. Use the definitions of kinetic energy, and voltage. You may need to reference these from previous tutorials. 1V=1J/1C Thus, energy = voltage x charge Kinetic energy = ½ mv 2 The energy given to the electrons from the voltage source is conserved and turned into kinetic energy. Thus the energy from the voltage equals the kinetic energy of the electrons. Vq=mv 2 /2 Rearranging for velocity gives: v= (2Vq/m) v= 2(100v)(1.6x10-19 C)/9.11x10-31 kg Answer is v = m/s. (A) is the correct answer choice.

3 Question No. 3 of If an electron in n=3 orbit of hydrogen atom jumps down to n = 2 orbit, the amount of energy released is. Question #03 A ev B ev C ev D. 1.5 ev E. 0 ev A. Incorrect! When an electron jumps from the energy state n 3 =3 to n 2 =2 then the amount of energy released is proportional to [1/n 1 2 1/n 2 2 ] ev and not [1/n 1 1/n 2 2 ] ev. B. Incorrect! Improper substitution yields a wrong result the answer is double the correct value. C. Correct! When an electron jumps from the energy state n 3 =3 to n 2 =2 then the amount of energy released is 1.89 ev. D. Incorrect The energy states figure as squares in the reciprocal hence, incorrect. When an electron jumps from the energy state n 3 =3 to n 2 =2 then the amount of energy released is proportional to [1/n 1 2 1/n 2 2 ] ev and not [1/n 1 1/n 2 2 ] ev. When an electron jumps from the energy state n 3 =3 to n 2 =2 then the amount of energy released is given by the relation E 2 E 1 = [1/n 2 2 1/n 3 2 ] ev = -13.6[1/4 1/9] = -1.89eV The negative sign shows that energy is released as the electron goes from a higher energy level to a lower one. (C) is the correct answer choice.

4 Question No. 4 of Calculate the de-brogile wave length of an electron whose kinetic energy is 50eV, (h=6.62 x joule-sec m o =9.1 x Kg). Note: 1 Angstrom = m. Question #04 A Angstroms B Angstroms C Angstroms D. 5.1 Angstroms E. 1.0 Angstroms A. Incorrect! The de Broglie wavelength is not evaluated properly the answer is double the actual value λ is not equal to 2h / m v. B. Correct! The de-broglie wavelength of a particle moving with velocity v is given by λ= h / m v = 1.73 atomic units of length, angstroms. C. Incorrect! The de Broglie wavelength is not evaluated properly the answer is double the actual value λ is not equal to h / 2m v. D. Incorrect! The de Broglie wavelength is not evaluated properly the answer is double the actual value λ is not equal to 3h / m v. The de-broglie wavelength of a particle moving with velocity v is given by λ= h / m v = 1.73 atomic units of length, angstroms. First change the existing energy into Joules x10 J 18 50eVx = 8x10 J 1eV Next, use the definition of kinetic energy to solve for the velocity of that electron. KE = mv v = v = 2 /2 2KE/m (2(8x10 18 J)/9.11x10 31 kg = 4.2x10 6 m/s Finally, use the debroglie wavelength formula to get the wavelength. λ = h/mv λ = 6.63x10 Js/(9.11x10 kg x 4.2x10 m/s) = 1.73x10 m = 1.73 Angstroms (B) is the correct answer choice.

5 Question No. 5 of Calculate the de Brogile wavelength associated with the golf ball of 50 gm moving with a velocity of 20 m/sec Question #05 A x m B x m C x m D x m E x m A. Correct! This is an application of the de Broglie wavelength. We know that λ = h/mv == x / 0.05 x 20 m = X m. B. Incorrect! λ x m improper substitution of the units. C. Incorrect! λ h/2mv hence, x m is incorrect and substitution of the units. D. Incorrect! λ 2h/3mv hence, x m is incorrect. Check your calculation. This is an application of the de Broglie wavelength. We know that λ = h/mv == x / 0.05 x 20 m = X m. This is an application of the de Broglie wavelength. We know that λ = h/mv Planck s constant h = x J-s Mass of the golf ball m = 50gm = 0.05Kg velocity of the golf ball v= 20 m/s λ = h/m v = x / 0.05 x 20 m = x m (A) is the correct answer choice.

6 Question No. 6 of What is the wavelength of light emitted when an electron transitions from the first excited state to the ground state in a hydrogen atom? Question #06 (A) 90 nm (B) 100 nm (C) 120 nm (D) 660 nm (E) 150 nm A. Incorrect. Do not use E = ev. B. Incorrect. Do not use E = (-13.6 ev) (1/3 2 1/1 2 ). C. Correct. Use E = (-13.6 ev) (1/2 2 1/1 2 ), and then use E = hc/λ to find the wavelength. D. Incorrect. Do not use E = (-13.6 ev) (1/3 2 1/2 2 ). E. Incorrect. Check your calculation. Use E = (-13.6 ev) (1/2 2 1/1 2 ), and then use E = hc/λ to find the wavelength. The difference in energy levels is given by: E= (-13.6 ev) - = 2 2 (-13.6 ev ) =10.2 ev The wavelength can be written in terms of the energy difference, Planck s constant, and the speed of light. Be sure to change the ev energy amount to into Joules with the proper conversion factor. hc E= λ ( J s ) ( m/s) hc λ= = -19 E 1.6x10 J 10. 2eV x 1eV = m = m =120 nm (C) is the correct answer choice.

7 Question No. 7 of If an X-ray photon is generated from an electron transition due a potential difference of 20 kv, what is the wavelength of the X-ray? Planck s constant, h = ev s. Question #07 (A) 0.06 Å (B) 0.6 Å (C) 6 Å (D) 60 Å (E) 600 Å A. Incorrect. Although this is a possible X-ray wavelength, it is one order of magnitude too small. B. Correct! Use Ε = hc/λ = qv to solve for the wavelength. C. Incorrect. Although this is a possible X-ray wavelength, it is one order of magnitude too large. D. Incorrect. Although this is a possible X-ray wavelength, it is two orders of magnitude too large. Use Ε = hc/λ = qv to solve for the wavelength. Relate the energy of a photon of wavelength λ to the energy due to accelerating an electron through a potential difference V: hc =qv λ The wavelength can be written in terms of the energy difference, Planck s constant, and the speed of light. hc λ= qv = ( ev s)( m/s) » 10 m»0.6 A 3 e V (B) is the correct answer choice.

8 Question No. 8 of Suppose that a laser produces light with a wavelength of 660 nm. If the laser delivers photons/s, what is the power output of the laser? Question #08 (A) 0.3 mw (B) 3 mw (C) 30 mw (D) 300 mw (E) 3 mw A. Incorrect. This answer is one order of magnitude too low. B. Correct! Use Ε = hc/λ, and P = re, where r is the rate of photons per second. C. Incorrect. This response is one order of magnitude too large. D. Incorrect. This response is two orders of magnitude too large. Use Ε = hc/λ, and P = re, where r is the rate of photons per second. Recall how the energy of a photon is related to its wavelength: hc E= λ. The power can be expressed as a product of rate (photons/s) and energy per photon: rhc P=rE= λ Finally, substitute the given numerical values and constants: ( s -1 ) ( J s ) ( m/s) P= m -3 =3 10 W =3 mw (B) is the correct answer choice.

9 Question No. 9 of Suppose that a 50.0 nm wavelength photon collides with and ionizes a hydrogen atom. If the atom was originally in the ground state, what is the kinetic energy of the ejected electron? Question #09 (A) 11.0 ev (B) 13.6 ev (C) 24.6 ev (D) 38.2 ev (E) 10.0 ev A. Correct. This is the remaining energy after the electron is ionized completely. B. Incorrect! This the minimum energy that a photon must have to become ionized. C. Incorrect. This is the energy of the 50.0 nm wavelength photon, not the kinetic energy of the ionized electron. D. Incorrect. Subtract, not add, the ground state energy from the energy of the photon to find the kinetic energy. Check your calculation. The kinetic energy can be found by subtracting the ground state energy from the energy of the photon. First, find the energy of the 50.0 nm wavelength photon: ( ev s ) ( m/s) hc E= = =24.6 ev -8 λ m The kinetic energy can be found by subtracting the ground state energy from the energy of the photon: K = ev ev = ev. The previous line shows how 13.6 ev are needed to ionize the electron. The remaining energy is left as kinetic energy of the electron. (A) is the correct answer choice.

10 Question No. 10 of What is the energy of an electromagnetic wave with a frequency of 10,000 Hz? Question #10 (A) J (B) J (C) J (D) J (E) J A. Incorrect. Do not divide Planck s constant by the frequency. B. Incorrect! This would be true if the frequency were 1 Hz. C. Correct! The energy of an electromagnetic wave is given by Planck s constant multiplied by the frequency. D. Incorrect. Do not multiply Planck s constant by the angular frequency. Check the exponent. It is too large. The energy of an electromagnetic wave is given by Planck s constant multiplied by the frequency. Use the equation for an energy, E, of an electromagnetic wave: E=hf Where h is Planck s constant, f is the frequency of the electromagnetic wave. Use the value of Planck s constant, and substitute the numerical value of the frequency given: ( ) E= J s ( Hz)= J. (C) is the correct answer choice.

1 The Cathode Rays experiment is associated. with: Millikan A B. Thomson. Townsend. Plank Compton

1 The Cathode Rays experiment is associated with: A B C D E Millikan Thomson Townsend Plank Compton 1 2 The electron charge was measured the first time in: A B C D E Cathode ray experiment Photoelectric