Question

${\huge{\tt{\orange{\fcolorbox{ref}{blue}{\underline{ QUESTION :}}}}}} \\ 1$


The longest wavelength doublet absorption transition is observed at 589 to 58.6 nm. Calculate the frequency of each transition and energy difference between two excited states.

Tʜɪꜱ Qᴜᴇꜱᴛɪᴏɴ Iꜱ Oɴʟʏ Fᴏʀ Mᴏᴅᴇʀᴀᴛᴏʀꜱ, Bʀᴀɪɴʟʏ Sᴛᴀʀꜱ Aɴᴅ Oᴛʜᴇʀ Bᴇꜱᴛ Uꜱᴇʀꜱ​

Answer 1

Step-by-step explanation:

Answer

The frequency of first transition is

ν

1

=

λ

1

c

=

589×10

−9

3.0×10

8

=5.093×10

14

s

−1

.

The frequency of second transition is

ν

2

=

λ

2

c

=

589.6×10

−9

3.0×10

8

=5.088×10

14

s

−1

.

The energy difference between two transitions is

ΔE=h(ν

2

−ν

1

)

=6.626×10

−34

×(5.093−5.088)×10

14

=3.313×10

−22

J.

hope it helps you✌

Answer 2

${\huge{\tt{\orange{\fcolorbox{ref}{blue}{\underline{AnSwEr:}}}}}}$

We know,

frequency = speed of light /wavelength

For wavelength ( λ₁ ) = 589 nm

frequency (v₁ ) = 3 × 10^8/589 × 10^-9

= 5.093 × 10¹⁴ Hz

For wavelength ( λ₂ ) = 589.6 nm

frequency ( v₂ ) = 3 × 10^8/589.6 × 10^-9

=5.088 × 10¹⁴ Hz

now, energy difference (∆E) = E₁ - E₂

∆E = E₁ - E₂ = hv₁ - hv₁

= h(v₁ - v₂)

= 6.626 × 10^-34 × ( 5.093 - 5.088) × 10¹⁴

= 6.626 × 0.005 × 10^(-34 + 14) J

= 3.31 × 10^-22 J