Two particles A1and A2 of masses m1 , m2 ( m1> m2) have the same de-Broglie
Then,
(a) their momenta are the same

(b) their energies are the same
(c) energy of A1 is less than the energy of A2
(d) energy of A1 is more than the energy of A2

Answer-(a,c)

2 Views|Posted 7 months ago
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1 Answer
A
7 months ago

This is a Multiple Choice Questions as classified in NCERT Exemplar

Explanation- as debroglie wavelength is λ = h m v = h p

P=h/ λ

P1/p2= λ 2 / λ 1

If wavelength are equal then ratio is 1:1 so P1=P2

E= 1/2mv2= p2/2m

E inversely proportional to m

So E 1 E 2 = m 2 m 1 <1

E12

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This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

f(x)=2x23x2x2=2x24x+x2x2=2x(x2)+1(x2)x2=(2x+1)(x2)x2=2x+1limx2f(x)=2x+1=limh02(2h)+1=4+1=5limx2+f(x)=2x+1=limh02(2+h)+1=4+1=5limx2f(x)=5Aslimx2f(x)=limx2+f(x)=limx2f(x)=5Hence,f(x)iscontinuousatx=2.

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Physics Ncert Solutions Class 12th 2023

Physics Ncert Solutions Class 12th 2023

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