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Problem Set 3: Bohr's Atom Solution

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Problem Set 3: Bohr's Atom Solution Modern Physics: Home work 3 Due date: 02 March 2015 Problem Set 3: Bohr's Atom Solution 1. A photon is emitted as an atom makes a transition from n = 4 to n = 2 level. What is the frequency, wavelength and energy of the emitted photon? Answer 1 we are given that, Initial orbit of electron = ni = 4 Final orbit of electron = nf = 2: Also we know that, Speed of electromagnetic waves = c = 3 × 108 m/s Planck's constant = h = 6:63 × 10−34 Js: We want to calculate the frequency, wavelength and energy of the emitted photon, when the atom makes a transition from ni = 4 to nf = 2. Frequency of the emitted photon can be calculated by using the equation, E E = hf; ) f = : h Let's at first calculate E by using the following relation, 1 1 E = −13:6 2 − 2 eV nf ni 1 1 = −13:6 − eV 22 42 1 1 = −13:6 − eV 4 16 3 = −13:6 × eV 16 = 2:55 eV: Thus frequency of photon will be, E 2:55 eV f = = h 6:63 × 10−34 Js 2:55 × 1:6 × 10−19 J = 6:63 × 10−34 Js = 6:15 × 1014 Hz: Spring semester 2015 1 Modern Physics: Home work 3 Due date: 02 March 2015 Wavelength of the emitted photon can be calculated by using the following equation. c = fλ c 3 × 108 m/s ) λ = = f 6:15 × 1014 Hz = 4:875 × 10−7 m = 488 nm: 2. For the Balmer series i.e., the atomic transitions where final state of the electron is n = 2, what is the longest and shortest wavelength possible? Is any of the frequency of Lyman series, which corresponds to transitions where electron ends up in n = 1 level, in the visible region? Answer 2 The atomic transitions where final state of the electron is n = 2, atoms emit a series of lines in the visible part of the spectrum. This series is called the Balmer Series. Balmer examined the four visible lines in the spectrum of the hydrogen atom. Wave length of this series can be calculated by using the equation, 1 1 1 = R − ; where n = 3; 4; 5; :::: λ 22 n2 R is the Rydberg constant, whose value is 1:097 × 107 m−1. The number n is just an integer; the above formula gives the longest wavelength, when n = 3, and gives each of the shorter wavelengths as n increases. From Balmer's equation, it looks like when n gets bigger, the lines should start getting really close together. That's exactly right; as n gets larger, 1 over n squared gets smaller, so there's less and less difference between the consecutive lines. We can see that the series has a limit, that is, as n gets larger and larger, the wavelength gets closer and closer to one particular value. If n is infinity, then 1 over n squared is 0, and the corresponding wavelength is shortest. The individual lines in the Balmer series are given the names Alpha, Beta, Gamma, and Delta, and each corresponds to a ni value of 3, 4, 5, and 6 respectively as shown in the figure below. 6 2 5 2 4 2 n=3 2 δ γ β α Spring semester 2015 2 Modern Physics: Home work 3 Due date: 02 March 2015 A picture from experimental demonstration is shown below. The longest wavelength corresponds to the smallest energy difference between energy levels, which in this case will be between n = 2 and n = 3. Wavelength for transition from n = 3 1 1 1 = R − λ 22 n2 1 1 = 1:097 × 107 m−1 × − 22 32 1 1 = 1:097 × 107 m−1 × − 4 9 5 = 1:097 × 107 m−1 × 36 = 1:524 × 106 m−1 ) λ = 6:56 × 10−7 m = 656 nm: Thus the longest wavelength in the Balmer series is 656 nm. The shortest wavelength corresponds to the largest energy difference between energy levels, which in this case will be between n = 2 and n = 1. Wavelength for transition from n = 1 1 1 1 = R − λ 22 n2 1 1 = 1:097 × 107 m−1 × − 22 12 1 1 = 1:097 × 107 m−1 × − 4 1 1 = 1:097 × 107 m−1 × 4 = 2:74 × 106 m−1 ) λ = 3:65 × 10−7 m = 365 nm: Spring semester 2015 3 Modern Physics: Home work 3 Due date: 02 March 2015 Thus the shortest wavelength in the Balmer series is 365 nm. In order to see that whether any of the frequency of Lymen series is in the visible region, let's calculate the largest and smallest frequencies of Laymen series. the wavelength of the Laymen series is given by, 1 1 = R 1 − ; where n = 2; 3; 4; 5; :::: λ n2 The longest wavelength corresponds to the smallest energy difference between energy levels, which in this case will be between n = 1 and n = 2. Wavelength for transition from n = 2 1 1 = R 1 − λ n2 1 = 1:097 × 107 m−1 × 1 − 22 1 = 1:097 × 107 m−1 × 1 − 4 3 = 1:097 × 107 m−1 × 4 = 8:23 × 106 m−1 ) λ = 1:22 × 10−7 m = 122 nm: Thus the longest wavelength in the Laymen series is 122 nm, and corresponding fre- quency will be, c 3 × 108 m/s f = = = 24:7 × 1014 Hz: λ 1:22 × 10−7 m The shortest wavelength corresponds to the largest energy difference between energy levels, which in this case will be between n = 1 and n = 1. Wavelength for transition from n = 1 1 1 = R 1 − λ n2 1 = 1:097 × 107 m−1 × 1 − 12 1 = 1:097 × 107 m−1 × 1 − 1 = 1:097 × 107 m−1 × 1 = 1:097 × 107 m−1 ) λ = 9:12 × 10−8 m = 91:2 nm: Spring semester 2015 4 Modern Physics: Home work 3 Due date: 02 March 2015 Thus the shortest wavelength in the Laymen series is 91:2 nm, and corresponding frequency will be, c 3 × 108 m/s f = = = 32:9 × 1014 Hz: λ 9:12 × 10−8 m From above we can see that frequency range of Laymen series is 24:7 × 1014 Hz to 32:9 × 1014. As frequency range of visible light is from 4 × 1014 to 8 × 1014, therefore no frequency of Laymen series is in the visible region. 3. A Hydrogen atom initially in its ground state i.e., n = 1 level, absorbs a photon and ends up in n = 4 level. What must have been the frequency of the photon? Now the electron makes spontaneous emission and comes back to the ground state. What are the possible frequencies of the photons emitted during this process. Answer 3 We are given that, Initial state of electron = ni = 1 final state of electron = nf = 4: We want to calculate the frequency of photon absorbed by the electron to make this transition. As we know that energy of photon absorbed is given by, 1 1 E = −13:6 2 − 2 eV nf ni 1 1 hf = −13:6 2 − 2 eV nf ni 13:6 1 1 f = − 2 − 2 eV h nf ni 13:6 1 1 = − −15 2 − 2 eV 4:14 × 10 eVs nf ni 15 1 1 = −3:28 × 10 2 − 2 Hz nf ni 1 1 f = −3:28 × 1015 − Hz 1!4 42 12 1 = −3:28 × 1015 − 1 Hz 16 −15 = −3:28 × 1015 × Hz 16 = +3:08 × 1015 Hz; Spring semester 2015 5 Modern Physics: Home work 3 Due date: 02 March 2015 where positive sign shows that photon has been absorbed. Spontaneous emission For spontaneous emission electron can have transitions as shown in the figure below. n=4 n=3 n=2 n=1 Direct Transition from n = 4 to n = 1 If electron jumps directly from level 4 to level 1, the frequency of the emitted photon will be equal to the frequency of the absorbed photon when it had jumped from level 1 to level 4, therefore, 15 f4!1 = −3:08 × 10 Hz; where negative sign shows that energy has been emitted. Indirect Transitions Indirectly the atom can make the following transitions to fall down from level 4 to level 1. (a) From level 4 to level 2 and then from 2 to level 1. (b) From level 4 to level 3 and then from 3 to level 1. (c) From level 4 to level 3, then from level 3 to level 2 and then finally from level 2 to level 1. Let's calculate frequencies for all these transitions one by one. Spring semester 2015 6 Modern Physics: Home work 3 Due date: 02 March 2015 From n = 4 to n = 2 1 1 f = −3:28 × 1015 − Hz 4!2 22 42 = −6:15 × 1014 Hz: From n = 2 to n = 1 1 1 f = −3:28 × 1015 − Hz 2!1 12 22 = −2:46 × 1015 Hz: From n = 4 to n = 3 1 1 f = −3:28 × 1015 − Hz 4!3 32 42 = −1:6 × 1014 Hz: From n = 3 to n = 1 1 1 f = −3:28 × 1015 − Hz 3!1 12 32 = −2:92 × 1015 Hz: From n = 3 to n = 2 1 1 f = −3:28 × 1015 − Hz 3!2 22 32 = −4:57 × 101 Hz: Negative sign with each frequency represents that photon has been emitted. 4. Non-relativistic Nature of Electron Inside Bohr's Atom Using Bohrs quantization rule, derive a formula for electrons speed in the quantized Bohrs orbits.
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