1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
harkovskaia [24]
2 years ago
5

the spectral lines observed for hydrogen arise from transitions from excited states back to the n=2 principle quantum level. Cal

culate the wavelengths associated with the spectral transitions of the hydrogen atom from the n=6,5,4 and 3 to the n=2 level.
Chemistry
2 answers:
Sunny_sXe [5.5K]2 years ago
5 0

Rydberg formula is given by:

\frac{1}{\lambda } = R_{H}\times (\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} )

where, R_{H} = Rydberg  constant = 1.0973731568508 \times 10^{7} per metre

\lambda = wavelength

n_{1} and n_{2} are the level of transitions.

Now, for n_{1}= 2 and n_{2}= 6

\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{6^{2}} )

= 1.0973731568508 \times 10^{7} \times (\frac{1}{4}-\frac{1}{36} )

= 1.0973731568508 \times 10^{7} \times (0.25-0.0278 )

= 1.0973731568508 \times 10^{7} \times 0.23

= 0.2523958\times 10^{7}

\lambda = \frac{1}{0.2523958\times 10^{7}}

= 3.9620\times 10^{-7} m

= 396.20\times 10^{-9} m

= 396.20 nm

Now, for n_{1}= 2 and n_{2}= 5

\frac{1}{\lambda} = 1.0973731568508 \times  10^{7} \times (\frac{1}{2^{2}}-\frac{1}{5^{2}} )

= 1.0973731568508 \times 10^{7} \times (0.25-0.04 )

= 1.0973731568508 \times 10^{7} \times (0.21 )

= 0.230 \times  10^{7}

\lambda= \frac{1}{0.230 \times 10^{7}}

= 4.3478 \times 10^{-7} m

= 434.78\times 10^{-9} m

= 434.78 nm

Now, for n_{1}= 2 and n_{2}= 4

\frac{1}{\lambda} = 1.0973731568508 \times  10^{7} \times (\frac{1}{2^{2}}-\frac{1}{4^{2}} )

=  1.0973731568508 \times 10^{7} \times (0.25-0.0625 )

= 1.0973731568508 \times 10^{7} \times (0.1875 )

= 0.20575 \times 10^{7}

\lambda= \frac{1}{0.20575 \times 10^{7}}

= 4.8602 \times 10^{-7} m

= 486.02 \times 10^{-9} m

= 486.02 nm

Now, for n_{1}= 2 and n_{2}= 3

\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{3^{2}} )

=  1.0973731568508 \times 10^{7} \times (0.25-0.12 )

=  1.0973731568508 \times 10^{7} \times (0.13 )

= 0.1426585\times 10^{7}

\lambda= \frac{1}{0.1426585\times 10^{7}}

= 7.0097 \times 10^{-7} m

= 700.97 \times 10^{-9} m

= 700.97 nm



eimsori [14]2 years ago
5 0

The wavelengths of spectral line observed in hydrogen atom are,

The value of wavelength of first spectral line from n=6 to n=2 is \boxed{{\text{410}}{\text{.2 nm}}}.

The value of wavelength of second spectral line from n=5 to n=2 is \boxed{{\text{434}}{\text{.1 nm}}} .

The value of wavelength of third spectral line from n=4 to n=2 is \boxed{{\text{486}}{\text{.2 nm}}} .

The value of wavelength of fourth spectral line from n=3 to n=2 is \boxed{{\text{656}}{\text{.3 nm}}} .

Further explanation:

Concept:

According to the Rydberg equation, the wavelength of spectral line related with the transition values as follows:

\frac{1}{\lambda }=\left( {{{\text{R}}_{\text{H}}}}\right)\left({\frac{1}{{{{\left({{{\text{n}}_{\text{f}}}}\right)}^2}}}-\frac{1}{{{{\left({{{\text{n}}_{\text{i}}}}\right)}^2}}}}\right)                       …… (1)

Here, \lambda is the wavelength of spectral line, {{\text{R}}_{\text{H}}}  is the Rydberg constant that has the value 1.097\times{10^7}{\text{ }}{{\text{m}}^{-1}} , {{\text{n}}_{\text{i}}}  is the initial energy level of transition, and {{\text{n}}_{\text{f}}}  is the final energy level of transition.

Therefore, after rearrangement of equation (1) \lambda can be calculated as,

\lambda=\frac{1}{{\left({1.097\times {{10}^7}{\text{ }}{{\text{m}}^{-1}}}\right)\left( {\frac{1}{{{{\left({{{\text{n}}_{\text{f}}}}\right)}^2}}}-\frac{1}{{{{\left({{{\text{n}}_{\text{i}}}}\right)}^2}}}}\right)}}                       …… (2)

Solution:

Finding the wavelength of spectral lines in each transition.

1. For the first transition, from initial energy level n=6 to final energy level n=2.

\begin{aligned}{\lambda _1}&=\frac{1}{{\left( {1.097\times {{10}^7}{\text{ }}{{\text{m}}^{ - 1}}}\right)\left({\frac{1}{{{{\left( {\text{2}}\right)}^2}}}-\frac{1}{{{{\left({\text{6}}\right)}^2}}}}\right)}}\\&= 4.102\times {10^{ - 7}}{\text{ m}}\times\left({\frac{{1{\text{ nm}}}}{{{{10}^{ - 9}}{\text{ m}}}}}\right)\\&={\mathbf{410}}{\mathbf{.2 nm}}\\\end{aligned}

2. For the second transition, from initial energy level n=5 to final energy level n=2.

\begin{aligned}{\lambda _2}&=\frac{1}{{\left( {1.097\times {{10}^7}{\text{ }}{{\text{m}}^{ - 1}}}\right)\left({\frac{1}{{{{\left( {\text{2}}\right)}^2}}}-\frac{1}{{{{\left({\text{5}}\right)}^2}}}}\right)}}\\&= 4.341\times {10^{ - 7}}{\text{ m}}\times\left({\frac{{1{\text{ nm}}}}{{{{10}^{ - 9}}{\text{ m}}}}}\right)\\&={\mathbf{434}}{\mathbf{.1 nm}}\\\end{aligned}

3. For the third transition, from initial energy level n=4 to final energy level n=2.

\begin{aligned}{\lambda _3}&=\frac{1}{{\left( {1.097\times {{10}^7}{\text{ }}{{\text{m}}^{ - 1}}}\right)\left({\frac{1}{{{{\left( {\text{2}}\right)}^2}}}-\frac{1}{{{{\left({\text{4}}\right)}^2}}}}\right)}}\\&= 4.862\times {10^{ - 7}}{\text{ m}}\times\left({\frac{{1{\text{ nm}}}}{{{{10}^{ - 9}}{\text{ m}}}}}\right)\\&={\mathbf{486}}{\mathbf{.2 nm}}\\\end{aligned}

4. For the fourth transition, from initial energy level n=3 to final energy level n=2.

\begin{aligned}{\lambda _4}&=\frac{1}{{\left( {1.097\times {{10}^7}{\text{ }}{{\text{m}}^{ - 1}}}\right)\left({\frac{1}{{{{\left( {\text{2}}\right)}^2}}}-\frac{1}{{{{\left({\text{3}}\right)}^2}}}}\right)}}\\&= 6.563\times {10^{ - 7}}{\text{ m}}\times\left({\frac{{1{\text{ nm}}}}{{{{10}^{ - 9}}{\text{ m}}}}}\right)\\&={\mathbf{656}}{\mathbf{.3 nm}}\\\end{aligned}

Learn more:

1. Ranking of elements according to their first ionization energy.: <u>brainly.com/question/1550767 </u>

2. Chemical equation representing the first ionization energy for lithium.: <u>brainly.com/question/5880605 </u>

<u> </u>

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Atomic structure

Keywords: transition, hydrogen atom, energy difference, transition from n=6 to n=2, transition from n=5 to n=2, transition from n=4 to n=2, transition from n=3 to n=2, spectral lines, wavelength of spectral lines.

You might be interested in
What's the volume of one mole of an ideal gas at Standard Temperature and Pressure? Question 17 options: A) 11.2 L B) 22.4 L C)
mel-nik [20]

Answer:

=zero degrease, 1 atm)? so the volume of an ideal gas is 22.l/mol at STP this, 22l.4Lis probably the most remembered and least useful number in chemistry

4 0
2 years ago
Three fundamental types of organic molecules are proteins, carbohydrates, and lipids. What is the fourth type?
Alja [10]

Answer:

Nucleic acids

That make our genes.

5 0
3 years ago
Read 2 more answers
In the compound AbO3, the ratio of aluminum to<br> oxygen is
jasenka [17]
The compound is Al2O3. The ratio of aluminum to oxygen is 2:3.
3 0
3 years ago
The temperature of nitrogen is unknown when the gas occupies 100.0 mL at 99.10 kPa. If the gas is known to occupy 74.2 mL at 133
laiz [17]

Answer:

The unknown temperature is 304.7K

Explanation:

V1 = 100mL = 100*10^-3L

P1 = 99.10kPa = 99.10*10³Pa

V2 = 74.2mL = 74.2*10^-3L

P2 = 133.7kPa = 133.7*10³Pa

T2 = 305K

T1 = ?

From combined gas equation,

(P1 * V1) / T1 = (P2 * V2) / T2

Solving for T1,

T1 = (P1 * V1 * T2) / (P2 * V2)

T1 = (99.10*10³ * 100*10^-3 * 305) / (133.7*10³ * 74.2*10^-3)

T1 = 3022550 / 9920.54

T1 = 304.67K

T1 = 304.7K

3 0
3 years ago
The constant volume heat capacity of a gas can be measured by observing the decrease in temperature when it expands adiabaticall
miv72 [106K]

Answer:

The value is  C_p  = 42. 8 J/K\cdot mol

Explanation:

From the question we are told that  

     \gamma = \frac{C_p }{C_v}

The  initial volume of the  fluorocarbon gas is  V_1 = V

 The final  volume of the fluorocarbon gas isV_2 = 2V

  The initial  temperature of the fluorocarbon gas is  T_1  =  298.15 K

  The final  temperature of the fluorocarbon gas is T_2  =  248.44 K

   The initial  pressure is P_1  = 202.94\  kPa

    The final   pressure is  P_2  =  81.840\  kPa

Generally the equation for  adiabatically reversible expansion is mathematically represented as

       T_2 =  T_1  * [ \frac{V_1}{V_2} ]^{\frac{R}{C_v} }

Here R is the ideal gas constant with the value  

        R =  8.314\  J/K \cdot mol

So  

   248.44 =   298.15  * [ \frac{V}{2V} ]^{\frac{8.314}{C_v} }

=> C_v  =  31.54 J/K\cdot mol

Generally adiabatic reversible expansion can also be mathematically expressed as

    P_2 V_2^{\gamma} = P_1 V_1^{\gamma}

=>[ 81.840 *10^3] [2V]^{\gamma} = [202.94 *10^3] V^{\gamma}    

=>  2^{\gamma} =  2.56

=>    \gamma =  1.356

So

     \gamma  =  \frac{C_p}{C_v} \equiv  1.356 = \frac{C_p}{31.54}

=>    C_p  = 42. 8 J/K\cdot mol

3 0
3 years ago
Other questions:
  • Why is nature more influential than nurture?
    13·2 answers
  • Which of the following is not a type of fault?
    6·1 answer
  • How much heat energy is required to raise the temperature of 0.367 kg of copper from 23.0 C to 60.0 C ? The specific heat of cop
    13·1 answer
  • A physical change occur when a
    12·2 answers
  • When do you use 6.02*10^23?? Do you use it when going to atoms or molecules????
    8·1 answer
  • How many grams of sodium metal are needed to react completely with 25.8 liters of chlorine gas at 293 Kelvin and 1.30 atmosphere
    15·1 answer
  • Why is it true that particles in a rock lying on the group have kinetic energy and energy
    15·1 answer
  • Explain the conditions under which goal succession occurs.​
    13·1 answer
  • Match the number of neutrons to the correct isotope by studying the atomic number and mass number of each isotope. Drag the item
    9·1 answer
  • Which of the following are common decomposers? (Select all that apply) fungi, ferns, bacteria, viruses, grasses​
    13·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!