An X - ray machine has an accelerating potential difference of 25,000 volts. By calculation the shortest wavelength will be obtained as
`(h=6.62 xx10^(-34)J-sec,e=1.6xx10^(-19)"coulomb")`

An X-ray machine has an accelerating potential difference of 25,000 volt. By calculating the shortest wavelength will be obtained as : ( h=6.62xx10^(-34)"J-sec", e=1.6xx10^(-19) coulomb)

An X-ray tube operates on 30 kV. What is the minimum wavelength emitted (h=6.6 xx10^(-34)Js,e=1.6xx10^(-19)"Coulomb",c=3xx10^(8)ms)

Calculate the energy of the photon of wave-length 4500A. Given h= 6.6xx10^(-34) J sec.

Under what potential difference should an electron be accelerated to obtain de Broglie, wavelength of 0.6overset@A ? (h=6.62xx10^(-34)j.s, m_e=9.1xx10^(-31) kg)

The energy of X-ray photon of wavelength 1.65 Å is (h=6.6 xx 10^(-34)J-sec,c=3xx10^(8)ms^(-1),1eV=1.6xx10^(-19)J)

What kV potential is to be applied on X-ray tube so that minimum wavelength of emitted X-ray may be 1 Å (h=6.625 xx 10^(-34) J-sec)

If a proton is accelerated through a potential difference of 1000V, then its de-Broglie wavelength is (given, m_(p) = 1.67 xx 10^(-27)kg, h = 6.63 xx 10^(-34)J-s )

A. What is de Broglie hypothesis? B. Write the formula for de Broglie wavelength. C. Calculate de Broglie wavelength associated with an electron accelerated by a potential difference of 100 volts. Given mass of the electron =9.1xx10^(-31)kg, h=6.634xx10^(-34)JS, 1 eV=1.6xx10^(-19)J

calculate the frequency of a photon of energy 3 eV. Givem h=6.62 xx10^(-34) Js.