If the velocity of the electron in Bohr's first orbit is 2.19x10 6 m/s . Calculate the de Broglie wavelength associated with it ?
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Answered by
4
momentum of electron =mass x velocity
=(9.1 x 10^-31)(2.19 x 10^6)
=19.929 x 10^-25kgm/s
now De-broglie wavelength,
wavelength =plank's constant/momentum
=(6.626 x 10^-34)/(19.929 x 10^-25)
=0.33 x10^-9 m
=(9.1 x 10^-31)(2.19 x 10^6)
=19.929 x 10^-25kgm/s
now De-broglie wavelength,
wavelength =plank's constant/momentum
=(6.626 x 10^-34)/(19.929 x 10^-25)
=0.33 x10^-9 m
Answered by
0
Answer:
v= 2.19×10^6 m/s
wavelength =h/mv
= 6.626 ×10^-34/ 9.1 ×10^-31× 2.19×10^6
= 0.3323×10^-9m
= 332.43pm
Explanation:
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