A hydrogen atom initially in the ground state absorbs a photon which excites it to the n = 4 level. Estimate the frequency of the photon. ( R = 1 . 097 × 10 7 m - 1 )

When the hydrogen atom is in the ground state, n 1 = 1 In the excited state, n 2 = 4 Wavelength ( λ ) can be calculated using the formula: 1 λ = R 1 ( n 1 ) 2 - 1 ( n 2 ) 2 1 λ = R 1 1 2 - 1 4 2 = 1 . 097 × 10 7 × 15 16 = 1 . 028 × 10 7 m - 1 The frequency of the photon, f = c λ Where c is the speed of light. f = 3 × 10 8 m / s × 1 . 028 × 10 7 m - 1 f = 3 . 09 × 10 15 Hz

State Bohr’s postulate to define stable orbits in hydrogen atom. How does de Broglie’s hypothesis explain the stability of these orbits ?

Bohr's postulate to define stable orbits in hydrogen atoms states that an electron can revolve only in those circular orbits in which its angular momentum is an integral multiple of n h / 2 π , where h is Planck’s constant and n is the principal quantum number of orbit. Thus, an electron can revolve only in certain orbits of definite radii. These are called stable orbits. An atom has a number of stable orbits in which an electron can reside without the emission of radiant energy. Each orbit corresponds, to a certain energy level. De Broglie hypothesis explanation of the stability of orbits: According to de Broglie, an electron orbiting around the nucleus is associated with a stationary wave. Electron wave is a circular standing wave. Since destructive interference will occur if a standing wave does not close upon itself, only those de Broglie waves exist for which the circumference of the circular orbit contains a whole number of wavelengths i.e., the circumference of n t h stable orbit for an electron can contain exactly n wavelength of de-Broglie. ⇒ n λ = 2 πr n n h p = 2 πr n n h m v = 2 πr n m v r n = n h 2 π ⇒ L = n h 2 π This was the quantum condition proposed by Bohr.

Calculate the orbital period of the electron in the first excited state of the hydrogen atom.

The radius of the n th orbit of the hydrogen and hydrogen like atoms is given by r n = 0 . 53 n 2 Z A ° where Z is atomic number. The velocity of the electron in n th orbit of hydrogen and hydrogen like atoms is given by v n = 2 . 2 × 10 6 × Z n m s - 1 The orbital period of the electron is given by T = 2 πr n v n Substituting the values from above, we get, T = 2 π × n × 0 . 53 × n 2 × 10 - 10 2 . 2 × 10 6 × Z × Z = 1 . 06 πn 3 × 10 - 10 2 . 2 × 10 6 × Z 2 For hydrogen atom, Z = 1 and for first excited state n = 2 . Put these values in above equation, we get T = 1 . 06 π 2 3 × 10 - 10 2 . 2 × 10 6 = 12 . 1 × 10 - 16 s . So, the orbital period of the electron in the first excited state is 1 . 21 × 10 - 15 s .

State Bohr ' s quantization condition of angular momentum. Calculate the shortest wavelength of the Bracket series and state to which part of the electromagnetic spectrum does it belong. ( Planck ' s constant = 6 . 6 × 10 - 34 J s , Rydberg ' s constant = 1 . 097 × 10 7 m - 1 ) .

According to Bohr, an electron in an atom could revolve in certain stable orbits without the emission of radiant energy and the electron revolves around the nucleus only in those orbits for which the angular momentum is some integral multiple of h 2 π where h is Planck's constant ( = 6 . 6 × 10 - 34 J s ) . The angular momentum L is given by L = mvr , where m is mass of electron, v is velocity in circular orbit, and R is radius of circular orbit. Therefore, L = mvr = nh 2 π where, n = 1 , 2 , 3 , . . . is called the principal quantum number and the above condition is known as the Bohr ' s quantization condition of angular momentum. Whenever an electron makes transition from one stationary orbit to another stationary orbit, energy is radiated in form of spectral line. The wavelength of a spectral line is given as 1 λ = RZ 1 n 1 2 - 1 n 2 2 where R is Rydberg ' s constant and Z is atomic number. The value of R is 1 . 097 × 10 7 m - 1 . For Bracket series, n 1 = 4 and n 2 = 5 , 6 , 7 , . . . Z = 1 for hydrogen atom and for the shortest wavelength of the Bracket series n 2 = ∞ . Substituting all the values in the above equation, we get, 1 λ = 1 . 097 × 10 7 1 4 2 - 1 ∞ 2 ⇒ 1 λ = 1 . 097 × 10 7 16 = 0 . 0681 × 10 7 m - 1 or, λ = 14 . 684 × 10 - 7 m or, λ = 1468 . 4 nm . Hence, the shortest wavelength of the Bracket series is 1468 . 4 nm . This wavelength belongs to infrared region of electromagnetic spectrum.