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**Introductory Exercise 30.1**

**Ques 1: The wavelength for n = 3 to n = 2 transition of the hydrogen atom is 656.3 nm. What are the wavelengths for this same transition in (a) positronium, which consists of an electron and a positron (b) singly ionized helium (Note: A positron is a positively charged electron).Sol:** (a) Reduced mass of positronium and electron is m/2 where m = mass of electron)

m has become half, so λ will become two times or 1312 nm or 1.31 nm.

For singly ionized helium atom z = 2

∴

Sol:

(c) An electron typically spends about 10

Sol:

∴

(b) ΔE =E

= 6.47 × 10

(c) In option (a), we have found that,

(a) A proton captures a μ

(b) Find the ionization energy of the atom.

Note Attempt this question after reading the whole chapter.

Sol:

= 2.55 × 10

(b) E ∝ m

∴ Ionization energy of given atom =

(m) (ionization energy of hydrogen atom)

= (207) (13.6 eV) = 2815.2 eV= 2.81 keV

(a) a 46 g golf ball with a velocity of 30 m/s,

(b) an electron with a velocity of 10

Sol:

(b) What if the electrons were replaced by photons of same energy?

Sol:

∴

Solving we get, n = 3.51

Hence electron jumps to n = 3.

So possible lines are between n = 3 to n = 2, n = 3 to n = 1 and between n = 2 to n = 1.

For n = 3 to n = 2

∴

Similarly other wavelengths can also be obtained.

(b) n = 3.51 (in option-a)

A photon always transfers its energy completely.

So it cannot excite the ground state electrons to n = 3 (like and electrons excited it in part-a).

Sol:

Sol:

E

= 1964 eV

∴ E

= -4613+1964 = -2649 eV

Sol:

⇒ ...(i)

Between 5E and 4E

5E - 4E = hf

∴ (from eq. (i))

Between 4E and E

4E - E = hf

∴

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