<span>Energy is calculated by molecule dividing energy by mole by Avogadro's number (6.022*10^23)
941kJ=9.41*10^5 J
so energy by molecule
E= 9.41*10^5/6.022*10^23=1.563*10^-18 J
Wavelength (w) given by E=hc/w
where, E = energy
h = planks constant (6.6262 x 10-34 J·s)
c = speed of light (3 x 10^8 m/s )
So,
w= hc/E
= (6.6262*10^-34)*(3*10^8) /1.563*10^-18
= 127.2 Nm
Longest wavelength of radiation =127.2 Nm</span>
Explanation:
According to Rydberg's formula, the wavelength of the balmer series is given by:
R is Rydberg constant for an especific hydrogen-like atom, we may calculate R for hydrogen and deuterium atoms from:
Here, 
is the "general" Rydberg constant, 
is electron's mass and M is the mass of the atom nucleus
For hydrogen, we have, 
:
Now, we calculate the wavelength for hydrogen:
![\frac{1}{\lambda}=R_H(\frac{1}{2^2}-\frac{1}{3^2})\\\lambda=[R_H(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=[1.0967*10^7m^{-1}(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=6.5646*10^{-7}m=656.46nm](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5Clambda%7D%3DR_H%28%5Cfrac%7B1%7D%7B2%5E2%7D-%5Cfrac%7B1%7D%7B3%5E2%7D%29%5C%5C%5Clambda%3D%5BR_H%28%5Cfrac%7B1%7D%7B2%5E2%7D-%5Cfrac%7B1%7D%7B3%5E2%7D%29%5D%5E%7B-1%7D%5C%5C%5Clambda%3D%5B1.0967%2A10%5E7m%5E%7B-1%7D%28%5Cfrac%7B1%7D%7B2%5E2%7D-%5Cfrac%7B1%7D%7B3%5E2%7D%29%5D%5E%7B-1%7D%5C%5C%5Clambda%3D6.5646%2A10%5E%7B-7%7Dm%3D656.46nm)
For deuterium, we have 
:
![R_D=\frac{1.09737*10^7m^{-1}}{(1+\frac{9.11*10^{-31}kg}{2*1.67*10^{-27}kg})}\\R_D=1.09707*10^7m^{-1}\\\\\lambda=[R_D(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=[1.09707*10^7m^{-1}(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=6.5629*10^{-7}=656.29nm](https://tex.z-dn.net/?f=R_D%3D%5Cfrac%7B1.09737%2A10%5E7m%5E%7B-1%7D%7D%7B%281%2B%5Cfrac%7B9.11%2A10%5E%7B-31%7Dkg%7D%7B2%2A1.67%2A10%5E%7B-27%7Dkg%7D%29%7D%5C%5CR_D%3D1.09707%2A10%5E7m%5E%7B-1%7D%5C%5C%5C%5C%5Clambda%3D%5BR_D%28%5Cfrac%7B1%7D%7B2%5E2%7D-%5Cfrac%7B1%7D%7B3%5E2%7D%29%5D%5E%7B-1%7D%5C%5C%5Clambda%3D%5B1.09707%2A10%5E7m%5E%7B-1%7D%28%5Cfrac%7B1%7D%7B2%5E2%7D-%5Cfrac%7B1%7D%7B3%5E2%7D%29%5D%5E%7B-1%7D%5C%5C%5Clambda%3D6.5629%2A10%5E%7B-7%7D%3D656.29nm)
I answered the question but it got deleted?? why?
Answer:
U = 30 m/s
a = 3 m/s²
Explanation:
"A car accelerating uniformly from rest reaches a maximum speed of U in 10 s. It then moves with that speed for an additional 20 s. The distance covered by the car in the 30 s interval is 750 m. Find U and the acceleration of the car in the first 10 s."
During the first 10 s:
v₀ = 0 m/s
v = U m/s
t = 10 s
The distance covered in this time is the average velocity times time:
Δx = ½ (v + v₀) t
Δx = ½ (U + 0) (10)
Δx = 5U
The distance covered in the next 20 seconds is speed times time:
Δx = 20U
The total distance is 750 m:
5U + 20U = 750
25U = 750
U = 30 m/s
The acceleration during the first 10 seconds is the change in speed over change in time.
a = Δv / Δt
a = (30 m/s − 0 m/s) / 10 s
a = 3 m/s²
Answer:
Hello
a) the change of velocity over time
