Question

The wavelength of maximum emitted energy of a body at 500 K is 3.16 pm . If the temperature of the body is raised by 500 K, the
wavelength of maximum emitted energy will be
(A) 2.16um
(B) 2.58um
(C) 6.32um
(D) 1.58 pm​

Answer 1

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Given :- }}}$

$\sf\implies Wavelength \ of \ maximum \ emmited \ energy \ of \ a \ body \ is\ 3.16 pm .\\\sf\implies The \ temperature \ is \ 500K$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: To \ Find :- }}}$

$\implies\textsf{ The wavelength of max. emitted energy} \\\qquad\textsf{if temperature is raised by 500 K.}$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Answer :- }}}$

If the temperature is raised by 500K , then the new temperature will be ( 500+500)K = 1000 K. So ,

$\underline{\boldsymbol{\purple{ According \ to \ Wien's \ Displacement \ Law :- }}}$

$\qquad\boxed{\green{\bf \lambda_{max} \propto \dfrac{1}{Temperature} }}$

$\sf:\implies \lambda_{max_1} T_1 = \lambda_{max_2} T_2 \\\\\sf:\implies 3.16 pm \times 500K = \lambda_{max_2} \times 1000K \\\\\sf:\implies \lambda_{max_2} = \dfrac{ 3.16 pm \times 500K}{1000K} \\\\\sf:\implies\boxed{\pink{\mathfrak{ \lambda_{max_2} = 1.58 \ picometer }}}$

$\underline{\blue{\sf Hence \ the \ required \ wavelength \ is \ \textsf{\textbf{1.58\ pm }}. }}$