[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. |
$\Rightarrow \text { For option (Q) } \mathrm{T}=3000$
$\lambda \mathrm{m}=\frac{\mathrm{b}}{\mathrm{T}}=\frac{2.9 \times 10^{-3}}{30000}$
$\lambda \mathrm{m}=\frac{2.9}{3} \times 10^{-6}$
$=0.96 \times 10^{-6}$
$=966.6 \mathrm{~nm}$
$\mathrm{P}_{3000}=6 \mathrm{~A}(3000)^{4}$
$\mathrm{P}_{6000}=6 \mathrm{~A}(6000)^{4}$
$\frac{\mathrm{P}_{3000}}{\mathrm{P}_{6000}}=\left(\frac{1}{2}\right)^4=\frac{1}{16}$
$\mathrm{P}_{3000}=\frac{1}{16} \mathrm{P}_{\text {क000 }}$
$\text { Q-4 }$
$\Rightarrow \text { For }(R) T=5000 \mathrm{~K}$
$\lambda \mathrm{m}=\frac{2.9 \times 10^{-3}}{5 \times 10^3}=0.58 \times 10^{-6}$
$=580 \mathrm{~mm}$
Visible to human eyes
$R-2$
$\Rightarrow$ For ($S$) $T=10,000 \rightarrow$ maximum
Hence $(3)$ is wrong as it has mmmmum ( $\lambda \mathrm{m})$
