Maximum speed of electron emitted from metal surface is `5xx10^(6 )ms^(-1)` .If specifi charge of electron is `1.8xx10^(11) ckg^(-1)`,then value of stopping potential will be…..

Maximum velocity of photoelectrons emitted by a metal surface is 1.2xx10^(6)m//s . Assuming the specific charge of the electrons to be 1.8xx10^(11)C//kg the value of stopping potential in volt will be:

Maximum velocity of the photoelectron emitted by a metal surface is 1.2 xx10^(6) m s^(-1) . Assuming the specific change of the electrons to be 1.8 xx10^(11) C kg ^(-1) kg the value of the stopping potential in volt will be

The charge of an electron is -1.6 xx10^(-19) C. The value of free charge on Li^+ ion will be

Maximum velocity of photoelectron emitted is 4.8 ms^(-1) . If e/m ratio of electron is 1.76xx10^(11)Ckg^(-1) , then stopping potential is given by

Maximum velocity of photo electrons is 3.5 xx 10^(6) m//s . If the specific charge of an electron is 1.75 xx 10^(11) C/kg, stopping potential of the electron is

An electron is accelerated in an electric field of 40V cm^(-1) . If e//m of electron is 1.76 xx10^(11) Ckg^(-1) , then its acceleration is

If the electronic charge e=1.6xx10^(-19)C and the specific charge e/m off electron is 1.76xx10^(11)C(kg)^(-1) then calculate the mass m of the electron.

The charge on the electron is 1.6 xx 10^(-19) C . The order of magnitude is

The charge of an electron is 1.6 xx 10^(-19)C what will be the value of charge on Na^(+) ion.