Question

The wave length of a microscope particle of mass 9.1times10^(-31) kg is 182nm its kinetic energy in Joules is​

Answer 1

Answer:

option b is the right one

Explanation:

debroglie wavelength /\ (lambda) is given by,

$(lambda) = \frac{h}{p} \\ = \frac{h}{ \sqrt{2mke} }$

where,

• h=Planck's constant
• p=momentum=√2mke
• m=mass of the particle
• ke=kinetic energy

therefore, kinetic energy, ke is given by,

$ke = \frac{ {h}^{2} }{2m {(lambda)}^{2} }$

substituting the given values in the above equation, we get,

$\frac{ {6.626 \times {10}^{ - 34} }^{2} }{2 \times 9.1 \times {10}^{ - 31} \times {182 \times {10}^{ - 9} }^{2} } \\ = 7.3 \times {10}^{ - 24}$

hope it helps...!!

Answer 2

SEE THE ATTACHMENT FOR DETAILED SOLUTION TO THE QUESTION DUDE ....❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤