The following MCQ on de Broglie waves appeared in Kerala Medical Entrance 2004 test paper:
An electron and a proton have the same de Broglie wave length. Then the kinetic energy of the electron is
(a) zero (b) infinity (c) equal to the kinetic energy of the proton (d) greater than the kinetic energy of the proton (e) none of these
Since the de Broglie wave length λ is given by λ = h/p where ‘h’ is Planck’s constant and ‘p’ is the linear momentum, we have p = h/λ. Therefore, the proton and the electron given in the question have the same linear momentum. The kinetic energy (E) is given by E = p2/2m. Since the mass (m) of the electron is less than that of the proton, it follows that the kinetic energy of the electron is greater [Option (d)].
Now consider the following MCQ:
A nucleus of mass ‘M’ at rest emits an α-particle of mass ‘m’. The de Broglie wave lengths of the α-particle and residual nucleus will be in the ratio
(a) m : M (b) (M+m) : m (c) M : m (d) √m:√M (e) 1 : 1
After the α-emission, the α-particle and the residual nucleus will fly off in opposite directions in accordance with the law of conservation of linear momentum. Since the momenta are equal in magnitude, the de Broglie wave lengths are the same and the ratio is 1:1 [Option (e)].
The following MCQ tests your basic knowledge regarding the electron orbits in the hydrogen atom:
The ratio of the de Broglie wave lengths of the electron in the first and the third orbits in the hydrogen atom is
(a) 1 : 1 (b) 1 : 3 (c) 1 : 9 (d) 1 : 6 (e) 1 : 27
You should remember that the orbit of quantum number ‘n’ is made of ‘n’ complete waves so that we have generally 2πrn = nλn where rnn is the radius of the nth orbit and λnn is the wave length of the electron in the nth orbit. So, we have 2πr1 = λ1 for the first orbit and 2πr3 = 3×λ3 for the third orbit. Therefore λ1/λ3 = 3r1/ r3.
But the orbital radius ‘r’ is directly proportional to n2. Therefore, λ1/λ3 = 3/9 = ⅓. The correct option is (b).
Now consider the following MCQ:
The kinetic energy of an electron is the same as that of a photon of wave length 3100 A.U. What is the wave length of this electron?
(a) 4 A.U. (b) 5.4 A.U. (c) 6.1 A.U. (d) 7.6 A.U. (e) 12.4 A.U.
In the case of a photon, the product λE = 12400 where the wave length λ is in Angstrom Unit (A.U.) and the energy E is in electron volt. Therefore, the energy of the photon of wave length 3100 A.U. is 12400/3100 = 4eV. Since the kinetic energy of the electron is the same as that of the photon, it follows that the kinetic energy of the electron is 4eV, which means this electron was accelerated by 4 volts. The wave length of an electron accelerated by ‘V’ volt is √(150/V) A.U. The answer to the problem is thus √(150/4) A.U. = √(37.5) = 6.1 A.U. approximately [Option (c)].
[You should remember that you can use the above simple relation for the wave length of an electron at small accelerating voltages only (in other words, at non-relativistic speeds only)].
An electron and a proton have the same de Broglie wave length. Then the kinetic energy of the electron is
(a) zero (b) infinity (c) equal to the kinetic energy of the proton (d) greater than the kinetic energy of the proton (e) none of these
Since the de Broglie wave length λ is given by λ = h/p where ‘h’ is Planck’s constant and ‘p’ is the linear momentum, we have p = h/λ. Therefore, the proton and the electron given in the question have the same linear momentum. The kinetic energy (E) is given by E = p2/2m. Since the mass (m) of the electron is less than that of the proton, it follows that the kinetic energy of the electron is greater [Option (d)].
Now consider the following MCQ:
A nucleus of mass ‘M’ at rest emits an α-particle of mass ‘m’. The de Broglie wave lengths of the α-particle and residual nucleus will be in the ratio
(a) m : M (b) (M+m) : m (c) M : m (d) √m:√M (e) 1 : 1
After the α-emission, the α-particle and the residual nucleus will fly off in opposite directions in accordance with the law of conservation of linear momentum. Since the momenta are equal in magnitude, the de Broglie wave lengths are the same and the ratio is 1:1 [Option (e)].
The following MCQ tests your basic knowledge regarding the electron orbits in the hydrogen atom:
The ratio of the de Broglie wave lengths of the electron in the first and the third orbits in the hydrogen atom is
(a) 1 : 1 (b) 1 : 3 (c) 1 : 9 (d) 1 : 6 (e) 1 : 27
You should remember that the orbit of quantum number ‘n’ is made of ‘n’ complete waves so that we have generally 2πrn = nλn where rnn is the radius of the nth orbit and λnn is the wave length of the electron in the nth orbit. So, we have 2πr1 = λ1 for the first orbit and 2πr3 = 3×λ3 for the third orbit. Therefore λ1/λ3 = 3r1/ r3.
But the orbital radius ‘r’ is directly proportional to n2. Therefore, λ1/λ3 = 3/9 = ⅓. The correct option is (b).
Now consider the following MCQ:
The kinetic energy of an electron is the same as that of a photon of wave length 3100 A.U. What is the wave length of this electron?
(a) 4 A.U. (b) 5.4 A.U. (c) 6.1 A.U. (d) 7.6 A.U. (e) 12.4 A.U.
In the case of a photon, the product λE = 12400 where the wave length λ is in Angstrom Unit (A.U.) and the energy E is in electron volt. Therefore, the energy of the photon of wave length 3100 A.U. is 12400/3100 = 4eV. Since the kinetic energy of the electron is the same as that of the photon, it follows that the kinetic energy of the electron is 4eV, which means this electron was accelerated by 4 volts. The wave length of an electron accelerated by ‘V’ volt is √(150/V) A.U. The answer to the problem is thus √(150/4) A.U. = √(37.5) = 6.1 A.U. approximately [Option (c)].
[You should remember that you can use the above simple relation for the wave length of an electron at small accelerating voltages only (in other words, at non-relativistic speeds only)].
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