Respuesta :
Answer:- wavelength of the bullet is [tex]2.42*10^-^3^4m[/tex] .
Solution:- de-Broglie equation is:
[tex]\lambda =\frac{h}{mv}[/tex]
where, [tex][tex]\lambda[/tex] is the de-Broglie wavelength, h is planck's constant, m is the mass in kg and v is the velocity in meter per second.
The given mass of the bullet is 5.97 g. Let's convert it to kg:
since, 1 kg = 1000 g
[tex]5.97g(\frac{1kg}{1000g})[/tex]
= 0.00597 kg
velocity is given as 1025 miles per hour. Let's convert it to meter per second as:
[tex]\frac{1025miles}{hour}(\frac{1.609km}{1mile})(\frac{1000m}{1km})(\frac{1hour}{60min})(\frac{1min}{60s})[/tex]
= [tex]\frac{458.12m}{s}[/tex]
Value of planck's constant is [tex]6.626*10^-^3^4J.s[/tex] .
Let's plug in the values in the equation and solve it for the wavelength.
[tex]\lambda =\frac{6.626*10^-^3^4J.s}{0.00597kg*458.12m.s^-^1}[/tex]
[tex]\lambda =2.42*10^-^3^4m[/tex]
note:( [tex]1J=1kg.m^2.s^-^2[/tex] )
So, the wavelength of the bullet is [tex]2.42*10^-^3^4m[/tex] .
