2.2×10^(-31) kg de Broglie 112 Å

2.2×10^(-31) kg de Broglie 112 Å

 

 

A particle with mass 2.2×10^(-31) kg

has a de Broglie wavelength of 112 Å.

Determine the momentum and kinetic energy of the particle.

 

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λ = h / p

( 참고 https://ywpop.tistory.com/3539 )

 

p = h / λ

= (6.626×10^(-34) J•s) / (112×10^(-10) m)

= (6.626×10^(-34) kg•m^2/s) / (112×10^(-10) m)

= (6.626×10^(-34)) / (112×10^(-10))

= 5.916×10^(-26) kg•m/s

 

> 1 J = 1 kg•m^2/s^2

 

 

 

KE = 1/2 mv^2

= (mv)^2 / 2m

= p^2 / 2m

( 참고 https://ywpop.tistory.com/6493 )

 

= (5.916×10^(-26) kg•m/s)^2 / (2×2.2×10^(-31) kg)

= (5.916×10^(-26))^2 / (2×2.2×10^(-31))

= 7.954×10^(-21) kg•m^2/s^2

= 7.954×10^(-21) J

 

 

 

1 eV = 1.60217646×10^(-19) J 이므로,

 

(7.954×10^(-21) J) (1 eV / 1.60217646×10^(-19) J)

= (7.954×10^(-21)) (1 / (1.60217646×10^(-19)))

= 0.04964 eV

= 4.964×10^(-2) eV