Atomic number of Hydrogen like species
that have a wave length difference of 59.3
nm between first line of Balmer and first line
of Lyman series is?
Answers
Answer:-
Given:-
Difference between the wave length of species in which it's atomic number is similar to Hydrogen between first line of balmer series and first line of Lyman series = 59.3 nm
We know;
Where;
- λ is the wavelength
is the Rydberg's constant. It's value is 1.09 × 10⁷ / m.
- n₁ & n₂ are ground and excited states of the electrons
- Z is the atomic number.
Also;
- The values of n₁ & n₂ are 1 & 2 in the first Lyman series.
- The values of n₁ & n₂ are 2 & 3 in the first balmer series.
Let , the wavelength in first Lyman series be λ₂ and wavelength in first balmer series be λ₁.
According to the question;
∴ The atomic number of the required element is 3.
Answer:
Answer:-
Given:-
Difference between the wave length of species in which it's atomic number is similar to Hydrogen between first line of balmer series and first line of Lyman series = 59.3 nm
We know;
\begin{gathered} \sf \: \dfrac{1}{ \lambda} = R_H \left| \dfrac{1}{ {n_1}^{2} }- \dfrac{1}{ {n_2}^{2} } \right| {Z}^{2} \\ \\ \\ \implies \sf \: \lambda = \frac{1}{R_H \left| \dfrac{1}{ {n_1}^{2} }- \dfrac{1}{ {n_2}^{2} } \right| {Z}^{2} } \end{gathered}
λ
1
=R
H
∣
∣
∣
∣
∣
n
1
2
1
−
n
2
2
1
∣
∣
∣
∣
∣
Z
2
⟹λ=
R
H
∣
∣
∣
∣
∣
n
1
2
1
−
n
2
2
1
∣
∣
∣
∣
∣
Z
2
1
Where;
λ is the wavelength
\sf R_HR
H
is the Rydberg's constant. It's value is 1.09 × 10⁷ / m.
n₁ & n₂ are ground and excited states of the electrons
Z is the atomic number.
Also;
The values of n₁ & n₂ are 1 & 2 in the first Lyman series.
The values of n₁ & n₂ are 2 & 3 in the first balmer series.
Let , the wavelength in first Lyman series be λ₂ and wavelength in first balmer series be λ₁.
According to the question;
\begin{gathered} \implies \sf \: \lambda_1 -\lambda_2 = 59.3 \: nm \\ \\ \\ \implies \sf \: \frac{1}{R_H \left| \dfrac{1}{ {2}^{2} }- \dfrac{1}{ {3}^{2} } \right| {Z}^{2} } - \frac{1}{R_H \left| \dfrac{1}{ {1}^{2} }- \dfrac{1}{ {2}^{2} } \right| {Z}^{2} } = 59.3 \times {10}^{ - 9} m \: \: \: \: ( 1 \: nm = {10}^{ - 9} \: m) \\ \\ \\ \implies \sf \: \frac{1}{R_H \times {Z}^{2} } \bigg( \frac{1}{ \frac{1}{4} - \frac{1}{9} } - \frac{1}{ \frac{1}{1} - \frac{1}{4} } \bigg) = 59.3 \times {10}^{ - 9} m \\ \\ \\ \implies \sf \: \frac{1}{ {Z}^{2} } \bigg( \frac{1}{ \frac{9 - 4}{36} } - \frac{1}{ \frac{4 - 1}{4} } \bigg) = 59.3 \times {10}^{ - 9} m \times R_H \\ \\ \\ \implies \sf \: \frac{1}{ {Z}^{2} } \bigg( \frac{ 36}{5} - \frac{4}{3} \bigg) = 59.3 \times {10}^{ - 9} m \times \frac{1.09 \times {10}^{7} }{m} \\ \\ \\ \implies \sf \: \frac{1}{ {Z}^{2} } \bigg( \frac{ 108 - 20}{15} \bigg) = 64.637 \times {10}^{ - 2} \\ \\ \\ \implies \sf \: \frac{1}{ {Z}^{2} } \bigg( \frac{ 88}{15} \bigg) = 64637 \times {10}^{ - 5} \\ \\ \\ \implies \sf \: \frac{1}{ {Z}^{2} } = 64637 \times \frac{1}{100000} \times \frac{ 15}{88} \\ \\ \\ \implies \sf \: {Z}^{2} = \frac{100000 \times 88}{64637 \times 15} \\ \\ \\ \\ \implies \sf \: {Z}^{2} \approx 9 \\ \\ \\ \implies \boxed{ \red{ \sf \:Z = 3}} \\ \\\end{gathered}
⟹λ
1
−λ
2
=59.3nm
⟹
R
H
∣
∣
∣
∣
∣
2
2
1
−
3
2
1
∣
∣
∣
∣
∣
Z
2
1
−
R
H
∣
∣
∣
∣
∣
1
2
1
−
2
2
1
∣
∣
∣
∣
∣
Z
2
1
=59.3×10
−9
m(1nm=10
−9
m)
⟹
R
H
×Z
2
1
(
4
1
−
9
1
1
−
1
1
−
4
1
1
)=59.3×10
−9
m
⟹
Z
2
1
(
36
9−4
1
−
4
4−1
1
)=59.3×10
−9
m×R
H
⟹
Z
2
1
(
5
36
−
3
4
)=59.3×10
−9
m×
m
1.09×10
7
⟹
Z
2
1
(
15
108−20
)=64.637×10
−2
⟹
Z
2
1
(
15
88
)=64637×10
−5
⟹
Z
2
1
=64637×
100000
1
×
88
15
⟹Z
2
=
64637×15
100000×88
⟹Z
2
≈9
⟹
Z=3
∴ The atomic number of the required element is 3.
Explanation:
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