Velocity of Electron ionized from Hydrogen

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Homework Help Overview

The discussion revolves around the kinetic energy of electrons that are ionized from hydrogen atoms when exposed to ultraviolet light with a wavelength of 45.0 nm. The original poster attempts to calculate the velocity of the freed electrons based on their kinetic energy.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of velocity from kinetic energy using the equation k = 0.5mv². The original poster expresses confusion over obtaining a velocity greater than the speed of light and questions whether a quantum equation is needed.

Discussion Status

Some participants provide guidance on unit conversion and suggest recalculating using kinetic energy in joules. There is acknowledgment of the importance of tracking units in calculations.

Contextual Notes

The original poster's calculations are based on kinetic energy expressed in electronvolts, which may lead to confusion when converting to standard units of velocity. The discussion highlights the need for careful attention to units in physics problems.

DRC12
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Homework Statement


Ultraviolet light with a wavelength of 45.0nm shines on a gas of hydrogen atoms in their ground states. Some of the atoms are ionized by the light. What is the kinetic energy of the electrons that are freed in this process?
I found the kinetic energy right but for some reason I thought I was looking velocity and when I solve for velocity I get a number much larger then the speed of light

Homework Equations


k=.5mv2
k=14eV
m=9.11*10-31kg

The Attempt at a Solution


k=.5mv2
v2=2k/m
v=√(2*14/9.11*10-31)=5.54*1015m/s
am I doing something wrong or is there some quantum equation needed
 
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DRC12 said:

Homework Statement


Ultraviolet light with a wavelength of 45.0nm shines on a gas of hydrogen atoms in their ground states. Some of the atoms are ionized by the light. What is the kinetic energy of the electrons that are freed in this process?
I found the kinetic energy right but for some reason I thought I was looking velocity and when I solve for velocity I get a number much larger then the speed of light

Homework Equations


k=.5mv2
k=14eV
m=9.11*10-31kg

The Attempt at a Solution


k=.5mv2
v2=2k/m
v=√(2*14/9.11*10-31)=5.54*1015m/s
am I doing something wrong or is there some quantum equation needed

Find the kinetic energy in joules, and calculate the speed with that value. Now you have the speed in units of eV0.5kg-0.5 instead of m/s.

ehild
 
Oh, thanks, I keep forgetting about that. I really need to start writing down the units
 
DRC12 said:
Oh, thanks, I keep forgetting about that. I really need to start writing down the units

It is a good idea !:smile:

ehild
 

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