Ask questions which are clear, concise and easy to understand.
Ask QuestionPosted by Aayush Singh 3 years, 6 months ago
- 1 answers
Posted by Rsrihari Haran 3 years, 6 months ago
- 2 answers
Ankush Phalswal 3 years, 6 months ago
Posted by Smruti Rekha Sethy 3 years, 6 months ago
- 1 answers
Rishabh Goswami 3 years, 6 months ago
Posted by 🤟Royal Thakur 🤟 2 years, 9 months ago
- 1 answers
Preeti Dabral 2 years, 9 months ago
A diode is a semiconductor device that essentially acts as a one-way switch for current. It allows current to flow easily in one direction, but severely restricts current from flowing in the opposite direction.
Posted by Akash Markandey 3 years, 6 months ago
- 0 answers
Posted by Bhojraj Singh 3 years, 6 months ago
- 1 answers
Ravi Kant Kumar 3 years, 6 months ago
Posted by Wangu Phago Wangu Phago 3 years, 7 months ago
- 0 answers
Posted by 🤟Royal Thakur 🤟 3 years, 7 months ago
- 5 answers
🤟Royal Thakur 🤟 3 years, 6 months ago
Posted by 🤟Royal Thakur 🤟 3 years, 7 months ago
- 3 answers
Posted by Arjina Pradhani 2 years, 9 months ago
- 1 answers
Preeti Dabral 2 years, 9 months ago
Here {tex}\lambda_{1}{/tex} = 6000 {tex}\overset \circ A{/tex}, {tex}\beta_{1}{/tex} = 0.8 mm,
{tex}\lambda_{2}{/tex} = 7500 {tex}\overset \circ A{/tex}
Fringe width in first case,
{tex}\beta_{1}=\frac{D \lambda_{1}}{d}{/tex}
Fringe width in second case,
{tex}\beta_{2}=\frac{D \lambda_{2}}{2 d}{/tex}
{tex}\therefore \frac{\beta_{2}}{\beta_{1}}=\frac{D \lambda_{2} / 2 d}{D \lambda_{1} / d}=\frac{1}{2} \cdot \frac{\lambda_{2}}{\lambda_{1}}{/tex}
or {tex}\beta_{2}=\frac{1}{2} \cdot \frac{\lambda_{2}}{\lambda_{1}} \cdot \beta_{1}{/tex}
= {tex}\frac{1}{2} \times \frac{7500 \overset \circ A}{6000 \overset \circ A} \times{/tex} 0.8 mm = 0.5 mm
Posted by Neha Mahato 3 years, 7 months ago
- 0 answers
Posted by Saurabh Kumar 3 years, 7 months ago
- 2 answers
Posted by 🤟Royal Thakur 🤟 3 years, 7 months ago
- 5 answers
🤟Royal Thakur 🤟 3 years, 7 months ago
🤟Royal Thakur 🤟 3 years, 7 months ago
Posted by Hitesh Tewatia 3 years, 7 months ago
- 0 answers
Posted by Parth Makhija 2 years, 9 months ago
- 1 answers
Preeti Dabral 2 years, 9 months ago
- Energy of the ground state (n = 1) = – (ionization energy) = –13.6 eV
The wavelength of the incident radiation, {tex}\lambda{/tex} = 975 {tex}\mathop {\text{A}}\limits^{\text{o}} {/tex}
{tex}\therefore{/tex} The energy of the incident photon = hc/{tex}\lambda{/tex}
= {tex}\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{975 \times 10^{-10} \times 1.6 \times 10^{-19}}{/tex} = 12.75 eV
Let electron is exerted to nth orbit,
{tex}\Rightarrow{/tex} 12.75 = 13.6 {tex}\left(\frac{1}{1^{2}}-\frac{1}{n^{2}}\right){/tex}
{tex}\Rightarrow{/tex} n = 4
The quantum transitions to the less excited states gives six possible lines as follows:
n = 4 : (4 {tex}\rightarrow{/tex} 3), (4 {tex}\rightarrow{/tex} 2), (4 {tex}\rightarrow{/tex} 1)
n = 3 : (3 {tex}\rightarrow{/tex} 2), (3 {tex}\rightarrow{/tex} 1)
n = 2 : (2 {tex}\rightarrow{/tex} 1)
- The longest wavelength emitted is for the transitions (4 {tex}\rightarrow{/tex} 3) where energy difference is minimum.
Emin = (E4 - E3) = 13.6 {tex}\left(\frac{1}{3^{2}}-\frac{1}{4^{2}}\right){/tex} = 0.661 eV
Thus {tex}\lambda_{\max }=\frac{h c}{\mathrm{E}_{\min }}{/tex}
= {tex}\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{0.661 \times 1.6 \times 10^{-19}} \mathrm{m}{/tex}
{tex}\approx{/tex} 18807 {tex}\mathop {\text{A}}\limits^{\text{o}} {/tex}
Posted by Lalit Mali 3 years, 7 months ago
- 2 answers
Shweta Sharma 3 years, 7 months ago
Riya Riya 3 years, 7 months ago
Posted by Mausam Kumar 2 years, 9 months ago
- 1 answers
Preeti Dabral 2 years, 9 months ago
Werner's theory states that- 1. Metals possess two types of valencies called primary / ionizable and secondary / non - ionizable valency. 2. Every metal atom has a tendency to satisfy both its primary and secondary valencies.
Posted by Suman Jais. 3 years, 7 months ago
- 1 answers
Devansh Chouhan 3 years, 7 months ago
Posted by ❤️Ritesh Gupta? 3 years, 7 months ago
- 3 answers
Siddhartha Raghava Raghava 3 years, 7 months ago
❤️Ritesh Gupta? 3 years, 7 months ago
Posted by ❤️Ritesh Gupta? 3 years, 7 months ago
- 5 answers
Pk . 3 years, 7 months ago
❤️Ritesh Gupta? 3 years, 7 months ago
Pk . 3 years, 7 months ago
Posted by Ranger King 3 years, 7 months ago
- 2 answers
Posted by Xoxo .. 3 years, 7 months ago
- 1 answers
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
Rajeev Goel 3 years, 6 months ago
0Thank You