de Broglie wavelength is
λ = h/p.
Since v is 0.1c, we have to use the relativistic linear momentum
(β=v/c = 0/1);
p = mₒ•β•c/sqrt(1- β²) =
= 9.1•10^-31•0.1•3•10^8/sqrt(1 – 0.1²) =
= 2.747•10^-23 kg/s.
λ = h/p =6.63•10^-34/2.747•10^-23 =
= 2.41•10^-11 m.
What is the de Broglie wavelength of an electron that strikes the back of the face of a TV screen at 1/10 the speed of light?
4 answers
Sorry for misprint
(β=v/c = 0.1)
(β=v/c = 0.1)
Thank you!! Also i was wondering if you could help me with this question as well
he energy difference between states A and B is twice the energy difference between states B and C (C > B > A). In a transition (quantum jump) from C to B, an electron emits a photon of wavelength 400 nm.
(a) What is the wavelength emitted when the photon jumps from B to A?
(b) What is the wavelength emitted when it jumps from C to A?
he energy difference between states A and B is twice the energy difference between states B and C (C > B > A). In a transition (quantum jump) from C to B, an electron emits a photon of wavelength 400 nm.
(a) What is the wavelength emitted when the photon jumps from B to A?
(b) What is the wavelength emitted when it jumps from C to A?
---------------C
↓ λ(CB)
---------------B
↓λ(BA)
---------------A
(a) ε(BA) = 2• ε(CB) =>
h•c/ λ(BA) =2• h•c/ λ(CB),
=> λ(BA) = λ(CB)/2 = 400/2 =200 nm.
(b) ε(CA) = ε(CB) + ε(BA) = ε(CB) + 2• ε(CB) = 3• ε(CB),
ε(CA) = 3• ε(CB), =>
λ(cA) = λ(CB)/3 = 400/3 =133.3 nm.
↓ λ(CB)
---------------B
↓λ(BA)
---------------A
(a) ε(BA) = 2• ε(CB) =>
h•c/ λ(BA) =2• h•c/ λ(CB),
=> λ(BA) = λ(CB)/2 = 400/2 =200 nm.
(b) ε(CA) = ε(CB) + ε(BA) = ε(CB) + 2• ε(CB) = 3• ε(CB),
ε(CA) = 3• ε(CB), =>
λ(cA) = λ(CB)/3 = 400/3 =133.3 nm.
1 answer