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6 (a) (i) The de Broglie equation is often written in the form $$\lambda = \frac{h}{mv}$$ Explain the meaning of each symbol used in the equation - OCR - A-Level Physics A - Question 6 - 2016 - Paper 1
Question 6
6 (a) (i) The de Broglie equation is often written in the form $$\lambda = \frac{h}{mv}$$ Explain the meaning of each symbol used in the equation. (ii) Describe b... show full transcript
Worked Solution & Example Answer:6 (a) (i) The de Broglie equation is often written in the form $$\lambda = \frac{h}{mv}$$ Explain the meaning of each symbol used in the equation - OCR - A-Level Physics A - Question 6 - 2016 - Paper 1
Step 1
Explain the meaning of each symbol used in the equation.
Answer
In the equation ( \lambda = \frac{h}{mv} ):
- ( \lambda ) is the de Broglie wavelength associated with a particle, which describes the wave-like behavior of particles.
- ( h ) is the Planck constant, a fundamental constant in quantum mechanics that relates energy and frequency of a photon.
- ( m ) is the mass of the particle, representing its inertia.
- ( v ) is the momentum of the particle, which is the product of its mass and velocity.
Step 2
Describe briefly one piece of evidence for believing that electrons sometimes behave like waves.
Answer
One piece of evidence is the phenomenon of electron diffraction. When electrons are passed through a thin sheet of graphite, they produce a pattern of rings on a fluorescent screen, indicating wave-like behavior due to interference patterns.
Step 3
Show that the final speed of an accelerated electron is about 1.3 x 10^8 ms^-1.
Answer
To find the speed of the electron, we can use the energy equation:
[ eV = \frac{1}{2} mv^2 ]
where ( e ) is the charge of the electron (approximately ( 1.6 \times 10^{-19} ) C) and ( V ) is the potential difference. Rearranging gives:
[ v = \sqrt{\frac{2eV}{m}} ]
Substituting ( e = 1.6 \times 10^{-19} , C ), ( V = 5.0 \times 10^4 , V ), and the mass of an electron ( m = 9.11 \times 10^{-31} , kg ):
[ v = \sqrt{\frac{2 \times (1.6 \times 10^{-19}) \times (5.0 \times 10^4)}{9.11 \times 10^{-31}}} \approx 1.3 \times 10^8 m/s ]
Step 4
Calculate the de Broglie wavelength of an electron moving at 1.3 x 10^8 ms^-1.
Answer
Using the de Broglie wavelength formula:
[ \lambda = \frac{h}{mv} ]
Substituting in the values:
- ( h = 6.63 \times 10^{-34} , Js )
- ( m = 9.11 \times 10^{-31} , kg )
- ( v = 1.3 \times 10^8 , m/s )
[ \lambda = \frac{6.63 \times 10^{-34}}{(9.11 \times 10^{-31}) \times (1.3 \times 10^8)} \approx 1.62 \times 10^{-13} , m ]
Step 5
Calculate by how many powers of 10 this is shorter than the wavelength of visible light used in optical microscopes.
Answer
The typical wavelength of visible light is about 500 nm or ( 5.0 \times 10^{-7} , m ).
To find how many powers of ten shorter the de Broglie wavelength is compared to the wavelength of visible light:
[ \frac{5.0 \times 10^{-7}}{1.62 \times 10^{-13}} = 3.09 \times 10^6 ]
This indicates that the de Broglie wavelength is 6 powers of ten shorter than that of visible light, as ( 3.09 \times 10^6 ) is roughly equivalent to ( 10^6 ).
Step 6
Name and describe briefly one piece of evidence which supports this particle-like behaviour.
Answer
One supporting evidence is the photoelectric effect, where light incident on a metal surface ejects electrons from it. The observations, such as the threshold frequency below which no electrons are emitted, can be explained using the photon model. Photons must transfer sufficient energy (at or above this threshold) to electrons to release them from the surface.