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Absorpstion spectrum of hydrogen shows line in 404.7 nm ; 435.8 nm and 486.3 nm. Determine:...

Absorpstion spectrum of hydrogen shows line in 404.7 nm ; 435.8 nm and 486.3 nm.

Determine:
a. Transtition for each line above
b. Ionization energy of Hydrogen

[Question doesn't state about spectral series (Lyman, Balmer.....). I am confused about this part. How I determine n1?]

Expert Solution

a) Given wavelengths fall in visible region. So the transition is related to Balmer series.

Relation between wavelength and corresponding states of transition is given by rydberg formula ,

1 / = R * [ 1 /n1² - 1/ n2² ] ............... ( 1 )

where R = 1.09677 * 10⁷ / m is Rydberg constant

and n1 , n2 are initial and final states of electron transition

( i ) Here n1 = 2 , as the transition corresponds to balmer series

wavelength given is = 404.7 nm = 404.7 × 10^-9 m

Then using Rydberg relation ,

1 / ( 404.7 × 10^-9 ) = 1.09677 × 10⁷ × [ 1/2² - 1 / n2² ]

0.00247 × 10⁹ = 1.09677 × 10⁷ × ( 0.25 - 1/n2² )

0.25 - 1/ n2² = ( 0.00247 × 10⁹ / 1.09677 × 10⁷ )

0.25 - 1/n2² = 0.225

1/n2² = 0.25 - 0.225

1/n2² = 0.025

n2² = 1 ÷ 0.025

n2 = 40

n2 = 6.32 = 6

So electron jumps from n1 = 2 to n2 = 6 to produce a line of wavelength 404.7 nm

( ii ) Wavelength = 435.8 nm

So we can write ,

1/ ( 435.8 × 10^-9 ) = 1.09677 × 10⁷ [ 1/4 - 1/n2² ]

( 0.00229 × 10⁹ ) / ( 1.09677 × 10⁷? ) = ( 0.25 - 1/n2² )

0.2092 = 0.25 - 1/n2²

1 /n2² = 0.25 - 0.2092

n2² = 1 / 0.041

n2 = 24.3

n2 = 5 ( approximately )

So a transition from n1 = 2 to n2 = 5 gives the wavelength 435.8 nm

( iii ) Wavelength = 486.3 nm = 486.3 × 10^-9 m

Using Rydberg formula again , we can write

1 / ( 486.3 × 10^-9 ) = 1.09677 × 10⁷ × [ 1/4 - 1/n2² ]

0.00205 × 10⁹ / ( 1.09677 × 10⁷ ) = ( 0.25 - 1/n2² )

0.1874 = 0.25 - 1/ n2²

1/n2² = 0.25 - 0.1874

n2² = 1 / 0.0626

n2 = 15.97

n2 = 4

( b ) When hydrogen atom is ionized , electron is said to move to a quantum state n2 = infinity from ground state n1 = 1

So we can write Rydberg equation as follows in this case

1/ = 1.09677 × 10⁷ × ( 1/ 1² - 1 / ² )

1/ = 1.09677 × 10⁷

= 1 / 1.09677 × 10⁷ = 0.911 × 10^-7 m

Corresponding energy E = h c /

= ( 6.626 × 10^-34 × 3 × 10⁸ ) / ( 0.911 × 10^-7? )

= 2.18 × 10^-19 J

= ( 2.18 × 10^-19 / 1.6 × 10^-19 ) eV

= 13.6 eV

So ionization energy of hydrogen is E = 13.6 eV

genius_generous answered 1 year ago

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