physics-class-12-unit-16-chapter-04-problem-solving-modern-physics-l-4-4-v7w8idxtupo

### <Success style="font-size:25px"> Problem Solving Modern Physics L-4 </Success>
### <primary style="font-size:24px"> Wave Nature of Particle </primary>
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- $\lambda = \frac{h}{p}$

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<footer style = "font-size: 18px">Problem-Solving Modern Physics $\rightarrow$ 20-th Century and Modern Physics $\rightarrow$ <span style = "color:blue"> Wave Nature of Particle </span> $\rightarrow$ Wave Nature of Particle $\rightarrow$ Theories </footer>

### <Success style="font-size:25px"> Problem Solving Modern Physics L-4 </Success>
### <primary style="font-size:24px"> Theories </primary>
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- **Max Plank (1900):** First breakthrough when Max Plank introduce the concept of Quantum of energy. The black body was explained in the way that energy exchanged between radiation and matter takes place in discrete/quantized way

- **Einstein Photoelectric effect (1905, Nobel prize in 1921):** Quantization of light in the form of photons

- **Bohr model (1911):** Combining the quantization of energy exchange and between the levels of atom

- **Compton (1923):** Confirm the particle nature of the light
 
 - **De Brogli (1923):** The wave nature of particle. So particle behave like wave

- **Davission and Germer (1927):** Interference pattern of electron beam from crystal

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![image](/subject-images/iitpal/image/physics-class-12-unit-16-chapter-04-problem-solving-modern-physics-l-4_4-v7w8idxtupo-04.jpg)

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<footer style = "font-size: 18px">Wave Nature of Particle $\rightarrow$ Wave Nature of Particle $\rightarrow$ <span style = "color:blue"> Theories </span> $\rightarrow$ Problem-1 $\rightarrow$ Solution-1 </footer>

### <Success style="font-size:25px"> Problem Solving Modern Physics L-4 </Success>
### <primary style="font-size:24px"> Problem-1 </primary>
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- In the historical experiment to determine Plank's constant, a metal surface was irradiated with the light of different wavelengths. The emitted photoelectrons energy were measured by applying a stopping potential. The relevant data for the wavelength $(\lambda)$ of incident light and the corresponding stopping potential (V) are given below. Given that $\mathrm{c}=3 \times 10^8 \mathrm{~m} / \mathrm{s}$, e $=1.6 \times 10^{-19} \mathrm{C}$ and Plank's constant (in units of $\mathrm{J} \mathrm{s}$ ) found from such experiment:

| $\lambda(\mu m)$ | V (Volt) | 
| --------         | -------- | 
| 0.3              | 2.0      | 
| 0.4              | 1.0      | 
| 0.5              | 0.4      |

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![image](/subject-images/iitpal/image/physics-class-12-unit-16-chapter-04-problem-solving-modern-physics-l-4_4-v7w8idxtupo-05.jpg)

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<footer style = "font-size: 18px">Wave Nature of Particle $\rightarrow$ Theories $\rightarrow$ <span style = "color:blue"> Problem-1 </span> $\rightarrow$ Solution-1 $\rightarrow$ Problem-2 </footer>

### <Success style="font-size:25px"> Problem Solving Modern Physics L-4 </Success>
### <primary style="font-size:24px"> Solution-1 </primary>
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- **Take any two Values.**

- $ \frac{h c}{\lambda}  =\phi+K E$

- $ k E  =\frac{h c}{\lambda}-\phi\rightarrow(1)$

- $2.0 \times 1.6 \times 10^{-19}  =\frac{h \times 3 \times 10^8}{0.3 \times 10^{-6}(\mathrm{~m})}-\phi \rightarrow(2)$

- $1.0 \times 1.6 \times 10^{-19}  =\frac{h \times 3.0 \times 10^8}{0.4 \times 10^{-6}(\mathrm{~m})}-\phi\rightarrow (3)$

- $(2) -(3)$

- $1.6 \times 10^{-19}  =\frac{h \times 3.0 \times 10^8}{10^{-7}}(\frac{1}{3}-\frac{1}{4})=\frac{h \times 3.0 \times 10^8}{10^{-7}}\left(\frac{1}{12}\right)$
 
- $h  =\frac{12 \times 10^{-7} \times 10^{-19} \times 1.6}{3.0 \times 10^8} $
- $h  =6.4 \times 10^{-34} \mathrm{~J} \cdot \mathrm{sec}$

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<footer style = "font-size: 18px">Theories $\rightarrow$ Problem-1 $\rightarrow$ <span style = "color:blue"> Solution-1 </span> $\rightarrow$ Problem-2 $\rightarrow$ Solution-2 </footer>

### <Success style="font-size:25px"> Problem Solving Modern Physics L-4 </Success>
### <primary style="font-size:24px"> Solution-6 </primary>
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- $\Rightarrow$ Incident Wavelength $(\lambda)=450 nm$

- $\text{Intensity} (I)=2.0 \mathrm{~W} / \mathrm{cm}^2$

- $=2.0 \mathrm{~J} / \mathrm{cm}^2 \cdot \mathrm{sec}$

* **$\text { No. of Photoelectron }$**

- $=45.24 \times 10^{17} \times \frac{5}{100}=2.263 \times 10^{17}$

- $\text{Current flew in outer circuit}$

- $=n \times e$

- $ =2.263 \times 10^{17} \times 1.6 \times 10^{-19}$

- $ =36 \mathrm{~mA}$

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<footer style = "font-size: 18px">Solution-5 $\rightarrow$ Problem-6 $\rightarrow$ <span style = "color:blue"> Solution-6 </span> $\rightarrow$ Problem-7 $\rightarrow$ Solution-7 </footer>

### <Success style="font-size:25px"> Problem Solving Modern Physics L-4 </Success>
### <primary style="font-size:24px"> Solution-7 </primary>
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- $k.\varepsilon_1 = 0.5 ~e V$

- $k.\varepsilon_2 = 1.2 ~e V$

- $\frac{h c}{\lambda_1}=\phi+k. \varepsilon_1-(1) \mid k. \varepsilon.=0.5 \mathrm{eV} $

- $ \frac{h c}{\lambda_2}=\phi+k. \varepsilon_2-(2) \mid k. \varepsilon.=1.2 \mathrm{eV} $

- $\phi=$ Work Function and it is the Material property

- **Using Equation-1**

- $\frac{h c}{532 \mathrm{~nm}}=\phi+0.5 \mathrm{ev}$

- $= \frac{6.63 \times 10^{-34} \times 3.0 \times 10^8}{532 \times 10^{-6 \mathrm{~m}}}$

- =$\phi+0.5 \times 1.6 \times 10^{-19} = 3.7 \times 10^{-19}$

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<footer style = "font-size: 18px">Solution-6 $\rightarrow$ Problem-7 $\rightarrow$ <span style = "color:blue"> Solution-7 </span> $\rightarrow$ Solution-7 $\rightarrow$ Problem-8 </footer>

### <Success style="font-size:25px"> Problem Solving Modern Physics L-4 </Success>
### <primary style="font-size:24px"> Solution-8 </primary>
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- $ h\nu=\phi+K E . $

- $\frac{h c}{\lambda_1}=\phi+k \cdot E_1-(1) $ 
- $\frac{h c}{\lambda_2}=\phi+K \cdot E_2-(2) $
- $  k \cdot E_1=\frac{1}{2} m v_1^2-(3) $
- $ k \cdot E_2=\frac{1}{2} m v_2^2-(4)$

- $ \frac{h c}{\lambda_1}-\phi=\frac{1}{2} m v_1^2-(5)$

- $ \frac{h c}{\lambda_2}-\phi=\frac{1}{2} m v_2^2-(6)$

- Divide (5/6)

- $\frac{v_1^2}{v_2{ }^2}=\frac{\frac{h c}{\lambda_1}-\phi}{\frac{h c}{\lambda_2}-\phi}-(7)$

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<footer style = "font-size: 18px">Solution-7 $\rightarrow$ Problem-8 $\rightarrow$ <span style = "color:blue"> Solution-8 </span> $\rightarrow$ Solution-8 $\rightarrow$ Problem-9 </footer>

### <Success style="font-size:25px"> Problem Solving Modern Physics L-4 </Success>
### <primary style="font-size:24px"> Solution-12 </primary>
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- $ w  =1.57 \times 10^7 \mathrm{C}$

- $f  =\frac{1.57 \times 10^7 \times 3.0 \times 10^8}{2 \times 3.14}\left(\frac{w}{2 \pi}\right) \cdot \mathrm{Hz}$

- $\phi=1.9 \mathrm{eV}$

- so
- $h \nu=  K \cdot E.+\phi $

- $(K . E .)_{e l e}  =h \nu-\phi $

- $=\frac{6.63 \times 10^{-34} \times 1.57 \times 10^7 \times 3.0 \times 10^8}{2 \times 3.14 \times 1.6 \times 10^{-19}} \mathrm{eV}-1.9 $

- $=(3.107-1.9) \mathrm{eV}$

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<footer style = "font-size: 18px">Solution-11 $\rightarrow$ Problem-12 $\rightarrow$ <span style = "color:blue"> Solution-12 </span> $\rightarrow$ Solution-12 $\rightarrow$ Problem-13 </footer>

### <Success style="font-size:25px"> Problem Solving Modern Physics L-4 </Success>
### <primary style="font-size:24px"> Solution-12 </primary>
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- $(K . E.)_{\text {electron }}=1.207 \mathrm{eV}$

- So stopping potential need to stop electron

- $V=1.207 \mathrm{~Volt}$

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<footer style = "font-size: 18px">Problem-12 $\rightarrow$ Solution-12 $\rightarrow$ <span style = "color:blue"> Solution-12 </span> $\rightarrow$ Problem-13 $\rightarrow$ Solution-13 </footer>

### <Success style="font-size:25px"> Problem Solving Modern Physics L-4 </Success>
### <primary style="font-size:24px"> Solution-14 </primary>
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- So $ V  =\frac{\sigma}{\epsilon_0} \times d=\frac{1 \times 10^{-9} \times 20}{8.85 \times 10^{-12} \times 100} $

- $ V =22.7 \mathrm{~V}$

- $\Rightarrow \frac{h c}{\lambda} =\phi+k \cdot E .$

- $ K E  =\frac{h c}{\lambda}-\phi-(1)$

- $ \frac{6.63 \times 10^{-34} \times 3.0 \times 10^8}{400 \times 10^{-9} \times 1.6 \times 10^{-19}}-1 . 9 e v $
- $ =3.105-19 $
- $K. E. =1.205$
- This $K . E.< < e V$ between the plates

- So min K. E. is $22.7 \mathrm{eV}$ and max. K.E. $=(22.7+1.205) \mathrm{eV}$

- $ K. E._{\text {max }}=23.905 \mathrm{eV} $

- $ K. E._{\text {min }}=22.7 \mathrm{e V}$

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<footer style = "font-size: 18px">Problem-14 $\rightarrow$ Solution-14 $\rightarrow$ <span style = "color:blue"> Solution-14 </span> $\rightarrow$ Problem-15 $\rightarrow$ Solution-15 </footer>

### <Success style="font-size:25px"> Problem Solving Modern Physics L-4 </Success>
### <primary style="font-size:24px"> Solution-15 </primary>
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- $\lambda  =400 \mathrm{~nm} $

- $\phi  =2.2 \mathrm{eV}$

- $E  =\frac{h C}{\lambda}=\frac{6.63 \times 10^{-34} \times 3.0 \times 10^8}{400 \times 10^{-9} \times 1.6 \times 10^{-19}}=3.1 \mathrm{eV}$

- After Collision, Energy lost.
- $\begin{aligned}& 3.1 \mathrm{ev} \times \frac{10}{100} \\& =0.31 \mathrm{ eV}\end{aligned}$

- So energy remains after $1^{\text {St }}$ Collision

- $3.1 \mathrm{ev}-0.31 \mathrm{ev}=2.7 \mathrm{9ev}$
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<footer style = "font-size: 18px">Solution-14 $\rightarrow$ Problem-15 $\rightarrow$ <span style = "color:blue"> Solution-15 </span> $\rightarrow$ Solution-15 $\rightarrow$ Solution-15 </footer>

### <Success style="font-size:25px"> Problem Solving Modern Physics L-4 </Success>
### <primary style="font-size:24px"> Solution-17 </primary>
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- Radiation in the range of $450 \mathrm{~nm}$ to $550 \mathrm{nm}$ Energy corresponding to $4.50 \mathrm{nm}$

- $\frac{1240 \text{ev nm}}{450 \mathrm{nm}}=2.75 \mathrm{eV}$

- For $550 \mathrm{~nm}$ :

- $\frac{1240 ~{ev. ~nm }}{550 nm} =2.26 \mathrm{eV}$

- The light come under visible range so transition from $n=2$

- $\text { to } n=3,4,5$

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<footer style = "font-size: 18px">Solution-16 $\rightarrow$ Problem-17 $\rightarrow$ <span style = "color:blue"> Solution-17 </span> $\rightarrow$ Solution-17 $\rightarrow$ Problem-18 </footer>

### <Success style="font-size:25px"> Problem Solving Modern Physics L-4 </Success>
### <primary style="font-size:24px"> Solution-20 </primary>
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- $\lambda  =\frac{h c}{E}=\frac{1240 \mathrm{ev} . \mathrm{nm}}{84 \times 10^2 \mathrm{~V}} =148 \mathrm{pm}$

- **For third Wave length**

- E  =$\frac{70}{100}(12-8.4) \times 10^3 =25.2 \times 10^2$

- $\lambda = \frac{hc}{E}= \frac{1240 {~ev}. nm}{25.2 \times 10^2}= 493 pm $

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<footer style = "font-size: 18px">Problem-20 $\rightarrow$ Solution-20 $\rightarrow$ <span style = "color:blue"> Solution-20 </span> $\rightarrow$ Thank You $\rightarrow$  </footer>