IIT JEE past question Thermodynamics 9

a black body is at temperature of 2880 K.

The energy of radiationemitted by this object with wavelength between 499 to 500 nm is U1, between 999 to 1000 nm is U2 and between 1499 and 1500 nm is U3.
The wien constant b=2.88x106 nm K.

Then
A) U1=0
B) U3=0
C) U1>U2
D) U2>U1

9 Answers

3
iitimcomin ·

2.88x10^6 = (2880)λ ...

10^3 = λ .....

thus U2>U1

1
skygirl ·

shudnt it be C?

λ=10E3

energy inv prop to wavelength.

so, U2 < U1...

pls correct if i am wrong......

1
ANKIT MAHATO ·

λT = b

λ where energy is max is b/T = 1000nm
so U2 is maximum as λ2 is close to it

so D) U1<U2

3
iitimcomin ·

I GUESS THE CURVE REACHES A MAXIMA FOR A PARTICULAR TEMP AND THEN FALLS DOWN AGAIN ............... SO THE MAX. ENGY. IS EMMITED IN THE WAVELENGTH 10^3 nm............

3
iitimcomin ·

nish bhiyya pls see if im rite ... confusion is in the air......

3
iitimcomin ·

3
iitimcomin ·

its either sky/mahto conceptual prob. or my conceptual prob.. only nish bhiyya can say.....[1]

1
ANKIT MAHATO ·

my conceptual error friend .. i edited my post !! g8 ... thanks for pointing out my error ![50]

1
skygirl ·

hmm.. u are correct... i din understand the question in the first place...

sorry for confusing...

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