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Chapter 6: Q41E (page 1356)

For the sodium atom of Example 40.8, find (a) the ground-state energy; (b) the wavelength of a photon emitted when the n= 4 to n= 3 transition occurs; (c) the energy difference for any 鈭�n=1transition.

Short Answer

Expert verified

(a) The ground state energy is 5.89脳10-3eV.

(b) The wavelength of photon when it makes a transition from n= 4 to n= 3 is 106渭m.

(c) The energy difference for any transition is0.0118 eV

Step by step solution

01

(a) Determination of the ground state energy.

For harmonic oscillator, the energy levels expression is,

En=n+12hk鈥�m=n+12h蝇

For ground state, n=0

So,

E0=12魔蝇=12hk鈥�m=1.055脳10鈭�34J脳s212.2N/m3.82脳10鈭�26kg=9.43脳10鈭�22J=5.89脳10鈭�3eV

Thus, the ground state energy is 5.89脳10-3eV.

02

(b) Determination of wavelength of photon when it makes a transition from n=4 to n=3.

For, n=4,

E4=4+12魔蝇=92魔蝇

For, n=3,

E3=3+12魔蝇=72魔蝇

Thus,

E4鈭�E3=92魔蝇鈭�72魔蝇=魔蝇=2E0=0.0118eV

Now, solve for the wavelength using equation,

位=hcE=6.63脳10鈭�34J鈰�s脳3.00脳108m/s1.88脳10鈭�21J=106渭m

Thus the wavelength emitted in transition is =106渭m.

03

(c) Determination of the energy difference for any ∆n=1 transition.

Like part (b), for n=卤1,鈥�i.e.鈥�螖n=1鈥� transition, the difference in energy level is,

En+1鈭�En=魔蝇=2E0=0.0118eV

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