To solve the problem, we need to determine the wavelength of a spectral line in the hydrogen spectrum that is caused by an electron transition between energy levels whose sum and difference are 5 and 3, respectively.
1. Understanding the Problem:
Given that the sum of the energy levels is 5 and the difference is 3, we can use the following relations for the energy levels involved in the transition: Let the two energy levels involved in the transition be \(n_1\) and \(n_2\). We are given the following: \[ n_1 + n_2 = 5 \quad \text{(sum of energy levels)} \] \[ n_1 - n_2 = 3 \quad \text{(difference of energy levels)} \] Solving these two equations simultaneously, we get: \[ n_1 = 4 \quad \text{and} \quad n_2 = 1 \] The wavelength of the spectral line for a hydrogen atom is given by the Rydberg formula: \[ \frac{1}{\lambda} = R \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right) \] Substituting the values of \(n_1 = 4\) and \(n_2 = 1\), we get: \[ \frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{4^2} \right) \] \[ \frac{1}{\lambda} = R \left( 1 - \frac{1}{16} \right) \] \[ \frac{1}{\lambda} = R \times \frac{15}{16} \] Thus, the wavelength \( \lambda \) is: \[ \lambda = \frac{16}{15R} \]
2. Identifying the Correct Answer:
The wavelength of the spectral line is \( \frac{16}{15R} \), which corresponds to Option A.
Final Answer:
The correct answer is Option A: \( \frac{16}{15R} \).
The elements of the 3d transition series are given as: Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zn. Answer the following:
Copper has an exceptionally positive \( E^\circ_{\text{M}^{2+}/\text{M}} \) value, why?
Match the Following
List-I (Use) | Item | Matches with | List-II (Substance) |
---|---|---|---|
A | Electrodes in batteries | II | Polyacetylene |
B | Welding of metals | III | Oxyacetylene |
C | Toys | I | Polypropylene |
An electrochemical cell is fueled by the combustion of butane at 1 bar and 298 K. Its cell potential is $ \frac{X}{F} \times 10^3 $ volts, where $ F $ is the Faraday constant. The value of $ X $ is ____.
Use: Standard Gibbs energies of formation at 298 K are:
$ \Delta_f G^\circ_{CO_2} = -394 \, \text{kJ mol}^{-1}; \quad \Delta_f G^\circ_{water} = -237 \, \text{kJ mol}^{-1}; \quad \Delta_f G^\circ_{butane} = -18 \, \text{kJ mol}^{-1} $
Assertion (A): Cu cannot liberate \( H_2 \) on reaction with dilute mineral acids.
Reason (R): Cu has positive electrode potential.
Consider the following electrochemical cell at standard condition. $$ \text{Au(s) | QH}_2\text{ | QH}_X(0.01 M) \, \text{| Ag(1M) | Ag(s) } \, E_{\text{cell}} = +0.4V $$ The couple QH/Q represents quinhydrone electrode, the half cell reaction is given below: $$ \text{QH}_2 \rightarrow \text{Q} + 2e^- + 2H^+ \, E^\circ_{\text{QH}/\text{Q}} = +0.7V $$
The percentage error in the measurement of mass and velocity are 3% and 4% respectively. The percentage error in the measurement of kinetic energy is: