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 $\frac{\mathrm{hc}}{\mathrm{e}}=1.24 \times 10^{-6} \mathrm{~V}-\mathrm{m}$ ]

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.
IIT 2023, Advanced
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$\Rightarrow$ For option $(\mathrm{P})$ temperature is minimum hence $\lambda \mathrm{m}$ will be maximum $\beta=\frac{\lambda \mathrm{D}}{\mathrm{d}} \Rightarrow \beta$ will also be maximum

$\Rightarrow \text { For option (Q) } \mathrm{T}=3000$

$\lambda \mathrm{m}=\frac{\mathrm{b}}{\mathrm{T}}=\frac{2.9 \times 10^{-3}}{30000}$

$\lambda \mathrm{m}=\frac{2.9}{3} \times 10^{-6}$

$=0.96 \times 10^{-6}$

$=966.6 \mathrm{~nm}$

$\mathrm{P}_{3000}=6 \mathrm{~A}(3000)^{4}$

$\mathrm{P}_{6000}=6 \mathrm{~A}(6000)^{4}$

$\frac{\mathrm{P}_{3000}}{\mathrm{P}_{6000}}=\left(\frac{1}{2}\right)^4=\frac{1}{16}$

$\mathrm{P}_{3000}=\frac{1}{16} \mathrm{P}_{\text {क000 }}$

$\text { Q-4 }$

$\Rightarrow \text { For }(R) T=5000 \mathrm{~K}$

$\lambda \mathrm{m}=\frac{2.9 \times 10^{-3}}{5 \times 10^3}=0.58 \times 10^{-6}$

$=580 \mathrm{~mm}$

Visible to human eyes

$R-2$

$\Rightarrow$ For ($S$) $T=10,000 \rightarrow$ maximum

Hence $(3)$ is wrong as it has mmmmum ( $\lambda \mathrm{m})$

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