Question

45. The kinetic energy (in J) be of a particle of mass
4.5x10-31 kg having a wavelength of 1000nm
is (h = 6.62 x 10-34 Js)
1) 2.43x10-24
2) 2.43x10-26
3) 4.86x10-24
4) 4.86x10-25


Answer is 4th option apparently how do we get it

Answer 1

Answer:

Answer is 4 number

Explanation:

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Answer 2

Concept:-

It might resemble a word or a number representation of the quantity's arithmetic value. It could resemble a word or a number that represents the numerical value of the quantity. It could have the appearance of a word or a number that denotes the quantity's numerical value.

Given:-

We have been given that "The kinetic energy (in J) be of a particle of mass $4.5\times 10^{-31}$ kg having a wavelength of $1000$nm is ($h = 6.62 \times 10^{-34}$ Js)"

Find:-

We need to find the option with the solution.

Solution:-

Applying de Broglie equation,

$v=\frac{h}{m \lamda}\\=\frac{6.62 \times 10^{-34}}{4.5 \times 10^{-31} \times 1000}$

Kinetic Energy of the given mass

$K.E=\frac{1}{2}mv^{2}\\=\frac{1}{2}4.5\times 10-31 \times {\frac{6.62 \times 10^{-34}}{4.5 \times 10^{-31} \times 1000}}^{2}\\=4.86\times 10^{-25}$

Hence, the statement "The kinetic energy (in J) be of a particle of mass $4.5\times 10^{-31}$ kg having a wavelength of $1000$ nm is ($h = 6.62 \times 10^{-34}$ Js)" is option $4) 4.86\times 10^{-25}$.

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