A particle wavelength (λp) of a proton is 10-12m.
What is the velocity of this proton [m/s]?
To achieved the velocity that you just computed, what must have been the potential difference (V) through which the proton was accelerated?
An electron of wavelength 1.74*10-10m strikes an atom of ionized helium (He+). What is the wavelength (m) of the light corresponding to the line in the emission spectrum with the smallest energy transition?
a) 3.96148749852*10^5
b) 820
c)....?
Thanks dear but actually I got the two first ones last night and stuck with part c. :(
Anyone for c) please?
what about some brainstorming???do we have to use Rydberg's equation??
What about doing your own work and not trying to cheat?
Please stop spamming.
To calculate the velocity of a proton, we can use the de Broglie wavelength equation:
λ = h / (m * v)
where λ is the wavelength, h is Planck's constant (6.63 * 10^-34 J·s), m is the mass of the particle, and v is the velocity.
Given that the proton's wavelength (λp) is 10^-12 m, we can rearrange the equation to solve for the velocity (v):
v = h / (m * λp)
Substituting the values into the equation:
v = (6.63 * 10^-34 J·s) / [(1.67 * 10^-27 kg) * (10^-12 m)]
Calculating this expression will give you the velocity of the proton in m/s.
To find the potential difference through which the proton was accelerated, we can use the equation for the kinetic energy of a charged particle accelerated through a potential difference:
K.E. = q * V
where K.E. is the kinetic energy, q is the charge of the proton (1.6 * 10^-19 C), and V is the potential difference.
Assuming the kinetic energy gained by the proton is equal to the energy difference between its initial state and final state, the potential difference (V) can be calculated by rearranging the equation:
V = K.E. / q
By substituting the known values into the equation, you will find the potential difference through which the proton was accelerated.
For the second question regarding the wavelength of light corresponding to the line in the emission spectrum with the smallest energy transition, we need to consider the Rydberg formula:
1/λ = R * (Z^2/nf^2 - Z^2/ni^2)
where λ is the wavelength, R is the Rydberg constant (approximately 1.097373 x 10^7 m^-1), Z is the atomic number (for helium, Z = 2), nf and ni are the final and initial energy levels, respectively.
Since we want to find the line with the smallest energy transition, we need to find the transition with the highest value of nf and the lowest value of ni. For helium (He+), we know that it has one electron missing, so nf = 2 (the second energy level). To find the lowest value of ni, we need to look at the energy levels of helium. Helium has two electrons, so the lowest available energy level would be when one electron is removed, giving us ni = 1 (the first energy level).
Substituting the values into the equation, we obtain:
1/λ = R * (2^2/2^2 - 2^2/1^2)
Simplifying this expression will give you the wavelength (λ) of the light corresponding to the line with the smallest energy transition.