Home
Class 11
CHEMISTRY
If the kinetic energy of an electron is ...

If the kinetic energy of an electron is `4.55 xx 10^(-25)J`, find its wavelength (Planck's cosntant, `h = 6.625 xx 10^(-34) kgm^(2)s^(-1), m = 9.1 xx 10^(-31)kg`).

Text Solution

AI Generated Solution

To find the wavelength of an electron given its kinetic energy, we can follow these steps: ### Step 1: Write down the given values - Kinetic Energy (KE) of the electron, \( KE = 4.55 \times 10^{-25} \, J \) - Planck's constant, \( h = 6.625 \times 10^{-34} \, kg \, m^2 \, s^{-1} \) - Mass of the electron, \( m = 9.1 \times 10^{-31} \, kg \) ### Step 2: Use the kinetic energy formula ...
Promotional Banner

Similar Questions

Explore conceptually related problems

Using Heisenberg's uncertainty principle, calculate the uncertainty in velocity of an electron if uncertainty in its position is 10^(-11)m Given, h =6.6 xx 10^(-14)kg m^2s^(-1), m=9.1 xx 10^(-31)kg

The energy of a photon is 3 xx10^(-12)erg .What is its wavelength in nm? (h=6.62 xx 10^(-27)erg-s,3 xx 10^(10)cm s^(-1))

The kinetic energy of an electron is 4.55 xx 10^(-25) J .The mass of electron is 9.1 xx 10^(-34) kg . Calculate velocity of the electron.

The kinetic energy of an electron is 4.55×10 ^ (−25) J. Calculate the wavelength, (h= 6.6×10 ^ (−34) Jsec,mass of electron= 9.1×10^ (−31) kg).

The work function of a certain metal is 3.31 xx 10^(-19) j then, the maximum kinetic energy of photoelectrons emitted by incident radiation of wavelength 5000 Å IS(GIVEN,h= 6.62 XX 10^(-34)J_(-s),c=3 xx 10^(8)ms^(-1),e=1.6 xx 10^(-19)C)

If de Broglie wavelength of an electron is 0.5467 Å, find the kinetic energy of electron in eV. Given h=6.6xx10^(-34) Js , e= 1.6xx10^(-19) C, m_e=9.11xx10^(-31) kg.

The uncertainty in momentum of an electron is 1 xx 10^-5 kg - m//s . The uncertainty in its position will be (h = 6.62 xx 10^-34 kg = m^2//s) .

The uncertainty in momentum of an electron is 1 xx 10^-5 kg - m//s . The uncertainty in its position will be (h = 6.62 xx 10^-34 kg = m^2//s) .

The uncertainty in momentum of an electron is 1 xx 10^-5 kg - m//s . The uncertainty in its position will be (h = 6.62 xx 10^-34 kg = m^2//s) .

If the mass of neutron is 1.7xx10^(-27) kg , then the de Broglie wavelength of neutron of energy 3eV is (h = 6.6xx10^(-34) J.s)