The kinetic energy of an electron is E when the incident wavelength is `lamda` To increase ti KE of the electron to 2E, the incident wavelength must be

The K.E of the electron is E when the incident wavelength is lamda . To increase the K.E of the electron to 2E, the incident wavelength must be

The KE of the photoelectrons is E when the incident wavelength is (lamda)/(2) . The KE becomes 2E when the incident wavelength is (lamda)/(3) . The work function of the metal is

The KE of the photoelectrons is E when the incident wavelength is (lamda)/(2) . The KE becomes 2E when the incident wavelength is (lamda)/(3) . The work function of the metal is

When the kinetic energy of an electron is increased , the wavelength of the associated wave will

When the kinetic energy of an electron is increased, the wavelength of the associated wave will

Maximum kinetic energy of a photoelectron is E when the wavelength of incident light is lambda . If energy becomes four times when wavelength is reduced to one thrid, then work function of the metal is

Maximum kinetic energy of a photoelectron is E when the wavelength of incident light is lambda . If energy becomes four times when wavelength is reduced to one thrid, then work function of the metal is

Kinetic energy of the particle is E and it's De-Broglie wavelength is lambda . On increasing it's KE by delta E , it's new De-Broglie wavelength becomes lambda/2 . Then delta E is

Kinetic energy of the particle is E and it's De-Broglie wavelength is lambda . On increasing it's KE by delta E , it's new De-Broglie wavelength becomes lambda/2 . Then delta E is