What potential difference is required in an electron microscope to give electrons a wavelength of 0.05 nm ?
For that de Broglie wavelength, Electron momentum = p = h/(5*10^-11 m)
where h is Planck's constant.
Momentum p = 6.62*10^-34/5*10^-11
= 1.32*10^-23 kg*m/s
Energy = p^2/(2m) = e*V
m = electron mass
e = electron charge
Solve for potential difference, V.
V = p^2/(2*m*e)
The de Broglie wavelength of an electron can be calculated using the equation:
λ = h / (m * v),
where λ is the wavelength, h is the Planck constant (6.626 x 10^-34 J s), m is the mass of the electron (9.109 x 10^-31 kg), and v is the velocity of the electron.
Since the electron microscope operates under high voltage, the kinetic energy of the accelerated electrons can be written as:
K.E. = (1/2) m * v^2.
The potential difference (V) is related to the kinetic energy (K.E.) of the electron by:
K.E. = e * V,
where e is the charge of the electron (1.602 x 10^-19 C).
By equating the two expressions for kinetic energy, we can solve for the velocity of the electron:
(1/2) m * v^2 = e * V,
v^2 = (2 * e * V) / m,
v = sqrt((2 * e * V) / m).
Now, substituting the expression for velocity into the de Broglie wavelength equation:
λ = h / (m * v) = h / (m * sqrt((2 * e * V) / m)),
Simplifying further, we get:
λ = h / sqrt(2 * e * V * m).
Rearranging the equation to solve for potential difference (V):
V = (h^2) / (2 * e * m * λ^2).
Plugging in the given values:
V = ((6.626 x 10^-34 J s)^2) / (2 * (1.602 x 10^-19 C) * (9.109 x 10^-31 kg) * (0.05 x 10^-9 m)^2),
V ≈ 571,000 volts.
Therefore, a potential difference of approximately 571,000 volts is required in an electron microscope to give the electrons a wavelength of 0.05 nm.
To determine the potential difference required in an electron microscope to give electrons a specific wavelength, we can use the de Broglie wavelength equation:
λ = h / √(2 * m * e * V)
Where:
λ is the wavelength of the electrons,
h is the Planck's constant (6.626 x 10^-34 J*s),
m is the mass of an electron (9.109 x 10^-31 kg),
e is the elementary charge (1.602 x 10^-19 C),
V is the potential difference.
We are given the wavelength (λ = 0.05 nm = 0.05 x 10^-9 m).
Rearranging the equation, we can isolate V:
V = (h^2) / (2 * m * e * λ^2)
Now, let's substitute the given values:
V = ((6.626 x 10^-34 J*s)^2) / (2 * (9.109 x 10^-31 kg) * (1.602 x 10^-19 C) * (0.05 x 10^-9 m)^2)
By plugging in the numbers and evaluating the expression, you can determine the potential difference required in the electron microscope to achieve a wavelength of 0.05 nm.