JEE Advance - Physics (2023 - Paper 1 Online - No. 15)
Match the temperature of a black body given in List-I with an appropriate statement in List-II, and choose the correct option.
[Given: Wien's constant as $2.9 \times 10^{-3} \mathrm{~m}-\mathrm{K}$ and $\frac{h c}{e}=1.24 \times 10^{-6} \mathrm{~V}-\mathrm{m}$ ]
[Given: Wien's constant as $2.9 \times 10^{-3} \mathrm{~m}-\mathrm{K}$ and $\frac{h c}{e}=1.24 \times 10^{-6} \mathrm{~V}-\mathrm{m}$ ]
| List - I | List - II |
|---|---|
| (P) $2000 \mathrm{~K}$ | (1) The radiation at peak wavelength can lead to emission of photoelectrons from a metal of work function $4 \mathrm{eV}$. |
| (Q) $3000 \mathrm{~K}$ | (2) The radiation at peak wavelength is visible to human eye. |
| (R) $5000 \mathrm{~K}$ | (3) The radiation at peak emission wavelength will result in the widest central maximum of a single slit diffraction. |
| (S) $10000 \mathrm{~K}$ | (4) The power emitted per unit area is $1 / 16$ of that emitted by a blackbody at temperature $6000 \mathrm{~K}$. |
| (5) The radiation at peak emission wavelength can be used to image human bones. |
$P \rightarrow 3, Q \rightarrow 5, R \rightarrow 2, S \rightarrow 3$
$P \rightarrow 3, Q \rightarrow 2, R \rightarrow 4, S \rightarrow 1$
$P \rightarrow 3, Q \rightarrow 4, R \rightarrow 2, S \rightarrow 1$
$P \rightarrow 1, Q \rightarrow 2, R \rightarrow 5, S \rightarrow 3$
Explanation
$\lambda_m T=b$
For option $(\mathrm{P})$ temperature is minimum hence $\lambda _\mathrm{m}$ will be maximum.
$\lambda_m=\frac{b}{T}=\frac{2.9 \times 10^{-3}}{2000}=1.45 \times 10^{-6} \mathrm{~m}=1450 \mathrm{~nm}$
$\begin{aligned} & \sin \theta=\frac{\lambda}{d} \\\\ & 2 \theta=\text { width of central maximum } \\\\ \Rightarrow & \text { width } \propto \lambda . \\\\ \Rightarrow & \text {maximum width of }(\mathrm{P}) \\\\ & \mathrm{P} \rightarrow 3\end{aligned}$
$\begin{aligned} & \Rightarrow \text { For option }(\mathrm{Q}),\mathrm{T}=3000 \\\\ & \\ & \lambda_\mathrm{m}=\frac{\mathrm{b}}{\mathrm{T}}=\frac{2.9 \times 10^{-3}}{30000} \\\\ & \lambda_\mathrm{m}=\frac{2.9}{3} \times 10^{-6} \\\\ & =0.96 \times 10^{-6} \\\\ & =966.6 \mathrm{~nm} \\\\ & \mathrm{P}_{3000}=6 \mathrm{~A}(3000)^4 \\\\ & \mathrm{P}_{6000}=6 \mathrm{~A}(6000)^4 \\\\ & \frac{\mathrm{P}_{3000}}{\mathrm{P}_{6000}}=\left(\frac{1}{2}\right)^4=\frac{1}{16} \\\\ & \mathrm{P}_{3000}=\frac{1}{16} \mathrm{P}_{6000} \\\\ & \mathrm{Q}-4\end{aligned}$
$\begin{aligned} \Rightarrow & \text { For }(\mathrm{R}), \mathrm{T}=5000 \mathrm{~K} \\\\ & \lambda_\mathrm{m}=\frac{2.9 \times 10^{-3}}{5 \times 10^3}=0.58 \times 10^{-6} \\\\ & =580 \mathrm{~nm} \\\\ & \text { Visible to human eyes } \\\\ & \mathrm{R}-2\end{aligned}$
For $(S), T=10,000 K$
$\lambda_m T=b \Rightarrow \lambda_m=\frac{2.9 \times 10^{-3}}{10,000}=290 \mathrm{~nm}$
$$ \begin{aligned} \phi & =4 e \mathrm{~V} \\\\ \Rightarrow \lambda_{\mathrm{T}} & =\frac{1240}{4}=310 \mathrm{n} \mathrm{m} \end{aligned} $$
To emit photoelectron
$$ \begin{aligned} & \lambda_0 < \lambda_{\mathrm{T}} \\\\ & \Rightarrow \mathrm{S} \rightarrow 1 \end{aligned} $$
For option $(\mathrm{P})$ temperature is minimum hence $\lambda _\mathrm{m}$ will be maximum.
$\lambda_m=\frac{b}{T}=\frac{2.9 \times 10^{-3}}{2000}=1.45 \times 10^{-6} \mathrm{~m}=1450 \mathrm{~nm}$
$\begin{aligned} & \sin \theta=\frac{\lambda}{d} \\\\ & 2 \theta=\text { width of central maximum } \\\\ \Rightarrow & \text { width } \propto \lambda . \\\\ \Rightarrow & \text {maximum width of }(\mathrm{P}) \\\\ & \mathrm{P} \rightarrow 3\end{aligned}$
$\begin{aligned} & \Rightarrow \text { For option }(\mathrm{Q}),\mathrm{T}=3000 \\\\ & \\ & \lambda_\mathrm{m}=\frac{\mathrm{b}}{\mathrm{T}}=\frac{2.9 \times 10^{-3}}{30000} \\\\ & \lambda_\mathrm{m}=\frac{2.9}{3} \times 10^{-6} \\\\ & =0.96 \times 10^{-6} \\\\ & =966.6 \mathrm{~nm} \\\\ & \mathrm{P}_{3000}=6 \mathrm{~A}(3000)^4 \\\\ & \mathrm{P}_{6000}=6 \mathrm{~A}(6000)^4 \\\\ & \frac{\mathrm{P}_{3000}}{\mathrm{P}_{6000}}=\left(\frac{1}{2}\right)^4=\frac{1}{16} \\\\ & \mathrm{P}_{3000}=\frac{1}{16} \mathrm{P}_{6000} \\\\ & \mathrm{Q}-4\end{aligned}$
$\begin{aligned} \Rightarrow & \text { For }(\mathrm{R}), \mathrm{T}=5000 \mathrm{~K} \\\\ & \lambda_\mathrm{m}=\frac{2.9 \times 10^{-3}}{5 \times 10^3}=0.58 \times 10^{-6} \\\\ & =580 \mathrm{~nm} \\\\ & \text { Visible to human eyes } \\\\ & \mathrm{R}-2\end{aligned}$
For $(S), T=10,000 K$
$\lambda_m T=b \Rightarrow \lambda_m=\frac{2.9 \times 10^{-3}}{10,000}=290 \mathrm{~nm}$
$$ \begin{aligned} \phi & =4 e \mathrm{~V} \\\\ \Rightarrow \lambda_{\mathrm{T}} & =\frac{1240}{4}=310 \mathrm{n} \mathrm{m} \end{aligned} $$
To emit photoelectron
$$ \begin{aligned} & \lambda_0 < \lambda_{\mathrm{T}} \\\\ & \Rightarrow \mathrm{S} \rightarrow 1 \end{aligned} $$
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