Problem 41E

A system releases 622 kJ of heat and does 105 kJ of work on the surroundings. What is the change in internal energy of the system?

Solution: The line spectrum observed in case of hydrogen atom can be quantitatively explained using Bohr’s model. When an electron jumps from a lower orbit to a higher orbit, it absorbs energy. When the electron jumps from higher orbit to lower orbit, it emits energy. The amount of energy absorbed or emitted when an electron jumps from orbit n(initial orbit) to i orbit n(final orbit) may be given as: f --------(1) Step1: In the Balmer series, for n = 3, the possible values of n = 4, 5, 6,.... Considering n = 3 and n = f i f i 4, the amount of energy emitted is given by: -18 1 1 E = (-2.18 x 10 J) ( 32 42 ) -18 1 1 = (-2.18 x 10 J) ( 9 16) = (-2.18 x 10 -18J)(7 / 144) = (-2.18 x 10 -18J) x (0.049) -18 = -0.107 x 10 J = -1.07 x 10 -19J Here, (-)ve sign indicates emission of energy. Step2: Wavelength of the radiation, is given by- = hc / E, where h is the Planck’s constant and c is the velocity of light -34 Here, Planck's constant, h = 6.626 x 10 J.s Velocity of light, c = 3 x 10 m/sec Substituting the value in the above expression, we get, = [(6.626 x 10 J.s) x (3 x 10 m/sec)] / (1.07 x 10 -19J) = 18.58 x 10 m -7 -6 = 1.858 x 10 m = 1858 x 10 m -9 Thus, wavelength of the radiation is 1858 nm. Step3: Out of all the radiations in the electromagnetic spectrum, the wavelength of the infrared radiation ranges from 7 x 10 m to 1 x 10 m. Thus, the electromagnetic radiation emitted lies in the infrared region and hence are not seen in the visible region of the spectrum. Hence, the correct option is (ii). Step4: b) In the Balmer series, for n = 2f nd n = 3,ihe amount of energy emitted is given by: E = (-2.18 x 10 -18J) 12 2 ( 2 3 ) = (-2.18 x 10 -18J) 1 1 ( 4 9) = (-2.18 x 10 -18J)(5 / 36) -18 = (-2.18 x 10 J) x (0.14) = -0.3052 x 10 -18J -19 = -3.05 x 10 J Here, (-)ve sign indicates emission of energy. Step5: Wavelength of the radiation, is given by- = hc / E, where h is the Planck’s constant and c is the velocity of light Here, Planck's constant, h = 6.626 x 10 J.s -34 8 Velocity of light, c = 3 x 10 m/sec Substituting the value in the above expression, we get, -34 8 -19 = [(6.626 x 10 J.s) x (3 x 10 m/sec)] / (3.05 x 10 J) = 6.517 x 10 m -7 = 651.7 x 10 m -9 Thus, wavelength of the radiation is 651.7 nm and this corresponds to red light of the spectrum. Step6: In the Balmer series, for n = 2, and n = 4, the amount of energy emitted is given by: f i E = (-2.18 x 10 -18J)( 12 2) 2 4 = (-2.18 x 10 -18J) ( 4 16 ) -18 = (-2.18 x 10 J)(12 / 64) = (-2.18 x 10 -18J) x (0.1875) -18 = -0.409 x 10 J = -4.09 x 10 -19J Here, (-)ve sign indicates emission of energy. Step7: Wavelength of the radiation, is given by- = hc / E, where h is the Planck’s constant and c is the velocity of light -34 Here, Planck's constant, h = 6.626 x 10 J.s Velocity of light, c = 3 x 10 m/sec Substituting the value in the above expression, we get, -34 8 -19 = [(6.626 x 10 J.s) x (3 x 10 m/sec)] / (4.09 x 10 J) = 4.86 x 10 m -7 -9 = 486 x 10 m Thus, wavelength of the radiation is 486 nm and this corresponds to blue-green light of the spectrum. Step8: In the Balmer series, for n = 2,f d n = 5, ti amount of energy emitted is given by: -18 1 1 E = (-2.18 x 10 J) ( 22 52) -18 1 1 = (-2.18 x 10 J) ( 4 25 ) = (-2.18 x 10 -18J)(21 / 100) -18 = (-2.18 x 10 J) x (0.21) = -0.4578 x 10 -18J = -4.578 x 10 -19J Here, (-)ve sign indicates emission of energy. Step9: Wavelength of the radiation, is given by- = hc / E, where h is the Planck’s constant and c is the velocity of light -34 Here, Planck's constant, h = 6.626 x 10 J.s Velocity of light, c = 3 x 10 m/sec Substituting the value in the above expression, we get, = [(6.626 x 10 J.s) x (3 x 10 m/sec)] / (4.578 x 10 -19J) -7 = 4.342 x 10 m = 434.2 x 10 m -9 Thus, wavelength of the radiation is 434.2 nm and this corresponds to blue-violet light of the spectrum. -------------------