Home
Class 11
CHEMISTRY
When a light of wavelength 300 nm falls ...

When a light of wavelength 300 nm falls on the surface of potassium metal, electrons with kinetic energy of `1.5 xx 10^(5) J mol^(-1)` are emitted, then calculate the minimum energy needed to remove an electron from the metal.

Promotional Banner

Topper's Solved these Questions

  • ATOMIC STRUCTURE

    AAKASH SERIES|Exercise SUBJECTIVE EXERCISE - 1 (VERY SHORT ANSWER QUESTIONS)|5 Videos
  • ATOMIC STRUCTURE

    AAKASH SERIES|Exercise SUBJECTIVE EXERCISE - 3 (VERY SHORT ANSWER QUESTIONS)|13 Videos
  • ATOMIC STRUCTURE

    AAKASH SERIES|Exercise SUBJECTIVE EXERCISE - 1 (LONG ANSWER QUESTIONS)|3 Videos
  • AROMATIC HYDROCARBONS

    AAKASH SERIES|Exercise OBJECTIVE EXERCIES - 3 (RECENT AIPMT/NEET QUESTIONS)|10 Videos
  • CHEMICAL BONDING

    AAKASH SERIES|Exercise OBJECTIVE EXERCISE -3 (RECENT AIPMT/NEET QUESTIONS )|39 Videos

Similar Questions

Explore conceptually related problems

When electromagnetic radiations of wave length 300nm falls on the surface of sodium, electrons are emitted with a kinetic energy of 1.68xx10^(5) J "mol"^(-1) . What is the minimum energy needed to remove an electron from sodium ? What is the maximum wavelength that will cause a photoelectron to be emitted ?

The energy required to emit an electron from the surface of a metal is called

The energy that is needed to remove an electron from the 1st Bohr orbit or Hydrogen atom is

Electrons with a kinetic energy of 6.023 xx 10^(4) J//mol are evolved from the surface of a metal, when it is exposed to radiation of wavelength of 600 nmn. The minimum amount of energy required to remove an electron from the metal atom is

Light of wavelength 5000^(@)A falls on a sensitive plate with photo electric work function 1.9eV. The kinetic energy of the photo electrons emitted will be

The ground state energy in J, of hydrogen atom is -X . The minimum energy in J, required to promote an electron from n=1 to n=2 in He^(+) is