Home
Class 12
PHYSICS
The potential energy of a particle of ma...

The potential energy of a particle of mass m is given by `U(x){{:(E_(0),,,0lexlt1),(0,,,xgt1):}`
`lambda_(1)` and `lambda_(2)` are the de-Broglie wavelength of the particle, when `0le x le1` and `x gt 1` respectively. If the total energy of particle is `2E_(0)`, then the ratio `(lambda_(1))/(lambda_(2))` will be

Promotional Banner

Similar Questions

Explore conceptually related problems

The potential energy of a particle of mass m is given by V(x) = lambda_(1) and lambda_(2) are the de - Brogle wavelength of the particle, when = lex le 1 and xgt 1 repectively ,if the total energy of particle is 2 E_(0) find lambda_(1) // lambda_(2)

The potential energy of particle of mass m varies as U(x)={(E_(0)"for"0lexle1),(0" for " gt 1):} The de Broglie wavelength of the particle in the range 0lexle1 " is " lamda_(1) and that in the range xgt1" is "lamda_(2) . If the total of the particle is 2E_(0)," find "lamda_(1)//lamda_(2) .

The potential energy of particle of mass m varies as U(x)={(E_(0)"for"0lexle1),(0" for " gt 1):} The de Broglie wavelength of the particle in the range 0lexle1 " is " lamda_(1) and that in the range xgt1" is "lamda_(2) . If the total of the particle is 2E_(0)," find "lamda_(1)//lamda_(2) .

The de Broglie wavelength (lambda) of a particle is related to its kinetic energy E as

lambda_(e),lambda_(p) and lambda_(alpha) are the de-Broglie wavelength of electron, proton and alpha particle. If all the accelerated by same potential, then

Velocity of particle = 4 v_e , if (lambda_p)/(lambda_e) = 2/1 . Find m_p/m_e ?