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 ~m-K and \frach ce=1.24 \times 10^-6 ~V-m ] .tg border-collapse:collapse;border-spacing:0;width:100% .tg tdborder-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal; .tg thborder-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal; .tg .tg-1wigfont-weight:bold;text-align:left;vertical-align:top .tg .tg-0laxtext-align:left;vertical-align:top List - I List - II (P) 2000 ~K (1) The radiation at peak wavelength can lead to emission of photoelectrons from a metal of work function 4 eV . (Q) 3000 ~K (2) The radiation at peak wavelength is visible to human eye. (R) 5000 ~K (3) The radiation at peak emission wavelength will result in the widest central maximum of a single slit diffraction. (S) 10000 ~K (4) The power emitted per unit area is 1 / 16 of that emitted by a blackbody at temperature 6000 ~K . (5) The radiation at peak emission wavelength can be used to image human bones.

P arrow 3, Q arrow 4, R arrow 2, S arrow 1

P arrow 3, Q arrow 5, R arrow 2, S arrow 3

P arrow 3, Q arrow 2, R arrow 4, S arrow 1

P arrow 1, Q arrow 2, R arrow 5, S arrow 3

Heat and Thermodynamics PYQ — JEE Advanced 2023 · Shift 1 · Q49 with Solution

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}

]

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.

A
$P \rightarrow 3, Q \rightarrow 5, R \rightarrow 2, S \rightarrow 3$
B
$P \rightarrow 3, Q \rightarrow 2, R \rightarrow 4, S \rightarrow 1$
C
$P \rightarrow 3, Q \rightarrow 4, R \rightarrow 2, S \rightarrow 1$
D
$P \rightarrow 1, Q \rightarrow 2, R \rightarrow 5, S \rightarrow 3$

View written solutionFree

Correct answer: C

Use Wien’s displacement law

For a black body,

\lambda_{\max} T = 2.9 \times 10^{-3}\ \text{m K}

So,

\lambda_{\max} = \frac{2.9\times 10^{-3}}{T}

Let us compute for each temperature.

(P) $T=2000\,\text{K}$
$\lambda_{\max} = \frac{2.9\times 10^{-3}}{2000}=1.45\times 10^{-6}\,\text{m}=1450\,\text{nm}$
This is in the infrared region.
(Q) $T=3000\,\text{K}$
$\lambda_{\max} = \frac{2.9\times 10^{-3}}{3000}=9.67\times 10^{-7}\,\text{m}=967\,\text{nm}$
This is also infrared.
(R) $T=5000\,\text{K}$
$\lambda_{\max} = \frac{2.9\times 10^{-3}}{5000}=5.8\times 10^{-7}\,\text{m}=580\,\text{nm}$
This lies in the visible region.
(S) $T=10000\,\text{K}$
$\lambda_{\max} = \frac{2.9\times 10^{-3}}{10000}=2.9\times 10^{-7}\,\text{m}=290\,\text{nm}$
This is ultraviolet.

Match with statement (3): widest central maximum in single slit diffraction

For a single slit,

\text{angular width of central maximum} \propto \lambda

So the largest wavelength gives the widest central maximum.

Among all temperatures, the largest $\lambda_{\max}$ is for 2000 K.

Hence,

P \rightarrow 3

Match with statement (4): power emitted per unit area is $1/16$ of that at $6000\,\text{K}$

Using Stefan–Boltzmann law,

E \propto T^4

We need

\left(\frac{T}{6000}\right)^4 = \frac{1}{16}

Taking fourth root,

\frac{T}{6000} = \frac{1}{2}

So,

T=3000\,\text{K}

Thus,

Q \rightarrow 4

Match with statement (2): radiation at peak wavelength is visible to human eye

Visible range is approximately

400\,\text{nm} \text{ to } 700\,\text{nm}

From above,

$1450\,\text{nm}$ : not visible
$967\,\text{nm}$ : not visible
$580\,\text{nm}$ : visible
$290\,\text{nm}$ : not visible

Hence,

R \rightarrow 2

Match with statement (1): peak radiation can emit photoelectrons from a metal of work function $4\,\text{eV}$

For photoelectric emission,

h\nu \ge \phi

\frac{hc}{\lambda} \ge 4\,\text{eV}

Given,

\frac{hc}{e}=1.24\times 10^{-6}\,\text{V m}

Thus photon energy in eV is

E = \frac{1.24\times 10^{-6}}{\lambda}\ \text{eV}

For (S), $\lambda=2.9\times 10^{-7}\,\text{m}$ :

E = \frac{1.24\times 10^{-6}}{2.9\times 10^{-7}} \approx 4.28\,\text{eV}

Since $4.28>4$ , photoemission is possible.

Hence,

S \rightarrow 1

Check statement (5): can be used to image human bones

Imaging bones requires X-rays, which have much smaller wavelengths than UV/visible/IR. None of the peak wavelengths except certainly not $5000\,\text{K}$ correspond to X-rays. So statement (5) is not used.

Final matching

Thus the correct matching is:

P \rightarrow 3,\quad Q \rightarrow 4,\quad R \rightarrow 2,\quad S \rightarrow 1

This corresponds to Option C.

More from Heat and Thermodynamics

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}

]

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.

A
$P \rightarrow 3, Q \rightarrow 5, R \rightarrow 2, S \rightarrow 3$
B
$P \rightarrow 3, Q \rightarrow 2, R \rightarrow 4, S \rightarrow 1$
C
$P \rightarrow 3, Q \rightarrow 4, R \rightarrow 2, S \rightarrow 1$
D
$P \rightarrow 1, Q \rightarrow 2, R \rightarrow 5, S \rightarrow 3$

View written solutionFree

Correct answer: C

Use Wien’s displacement law

For a black body,

\lambda_{\max} T = 2.9 \times 10^{-3}\ \text{m K}

So,

\lambda_{\max} = \frac{2.9\times 10^{-3}}{T}

Let us compute for each temperature.

(P) $T=2000\,\text{K}$
$\lambda_{\max} = \frac{2.9\times 10^{-3}}{2000}=1.45\times 10^{-6}\,\text{m}=1450\,\text{nm}$
This is in the infrared region.
(Q) $T=3000\,\text{K}$
$\lambda_{\max} = \frac{2.9\times 10^{-3}}{3000}=9.67\times 10^{-7}\,\text{m}=967\,\text{nm}$
This is also infrared.
(R) $T=5000\,\text{K}$
$\lambda_{\max} = \frac{2.9\times 10^{-3}}{5000}=5.8\times 10^{-7}\,\text{m}=580\,\text{nm}$
This lies in the visible region.
(S) $T=10000\,\text{K}$
$\lambda_{\max} = \frac{2.9\times 10^{-3}}{10000}=2.9\times 10^{-7}\,\text{m}=290\,\text{nm}$
This is ultraviolet.

Match with statement (3): widest central maximum in single slit diffraction

For a single slit,

\text{angular width of central maximum} \propto \lambda

So the largest wavelength gives the widest central maximum.

Among all temperatures, the largest $\lambda_{\max}$ is for 2000 K.

Hence,

P \rightarrow 3

Match with statement (4): power emitted per unit area is $1/16$ of that at $6000\,\text{K}$

Using Stefan–Boltzmann law,

E \propto T^4

We need

\left(\frac{T}{6000}\right)^4 = \frac{1}{16}

Taking fourth root,

\frac{T}{6000} = \frac{1}{2}

So,

T=3000\,\text{K}

Thus,

Q \rightarrow 4

Match with statement (2): radiation at peak wavelength is visible to human eye

Visible range is approximately

400\,\text{nm} \text{ to } 700\,\text{nm}

From above,

$1450\,\text{nm}$ : not visible
$967\,\text{nm}$ : not visible
$580\,\text{nm}$ : visible
$290\,\text{nm}$ : not visible

Hence,

R \rightarrow 2

Match with statement (1): peak radiation can emit photoelectrons from a metal of work function $4\,\text{eV}$

For photoelectric emission,

h\nu \ge \phi

\frac{hc}{\lambda} \ge 4\,\text{eV}

Given,

\frac{hc}{e}=1.24\times 10^{-6}\,\text{V m}

Thus photon energy in eV is

E = \frac{1.24\times 10^{-6}}{\lambda}\ \text{eV}

For (S), $\lambda=2.9\times 10^{-7}\,\text{m}$ :

E = \frac{1.24\times 10^{-6}}{2.9\times 10^{-7}} \approx 4.28\,\text{eV}

Since $4.28>4$ , photoemission is possible.

Hence,

S \rightarrow 1

Check statement (5): can be used to image human bones

Final matching

Thus the correct matching is:

P \rightarrow 3,\quad Q \rightarrow 4,\quad R \rightarrow 2,\quad S \rightarrow 1

This corresponds to Option C.

Heat and Thermodynamics question

Heat and Thermodynamics question

More from Heat and Thermodynamics

Heat and Thermodynamics question

Heat and Thermodynamics question

More from Heat and Thermodynamics