Show that the de Broglie wavelength of an electron accelerated through a potential difference of V volts is λ = h 2 m e V .

Kinetic energy of an electron accelerated through a potential V : K . E . e l e c t r o n = e V K . E . e l e c t r o n = 1 2 m e v e 2 = 1 2 p e 2 m e Here, e , m e , v e and p e are charge on electron, mass of electron, velocity of electron and momentum of electron respectively. p e = 2 m e K . E . e l e c t r o n = 2 m e e V Now, the de Broglie wavelength of electron: λ e = h p e = h 2 m e e V . Hence, the de Broglie wavelength of an electron accelerated through a potential difference of V volts is λ e = h 2 m e e V .

Define work function and threshold frequency of a metal. Write Einstein’s photoelectric equation in terms of stopping potential, frequency of incident radiation and threshold frequency of a metal.

Work function ϕ ∘ is the minimum amount of energy required to remove an electron to infinity, from the surface of the metal.(i.e. ϕ ∘ = h v ∘ ) Threshold frequency v ∘ is the critical frequency of incident light that cause ejection of electrons without giving them additional energy. So Einstein's photoelectric equation: h ν = ϕ ∘ + K . E . m a x ⇒ K . E . m a x = h ν - ϕ ∘ = h ν - v ∘ Here, V s is the stopping potential and K . E . m a x is the maximum kinetic energy of emitted photoelectron. Since, K . E . m a x = e V s . By substituting this in above equation: e V s = h ( v - v ∘ ) ⇒ v ∘ = v - e V s h . Hence, Einstein’s photoelectric equation in terms of stopping potential, frequency of incident radiation and threshold frequency of a metal is v ∘ = v - e V s h .

What is the effect of intensity of incident light on photoelectric current?

Photoelectric current flows when light of frequency greater than threshold frequency falls on photo-cathode. Then photoelectrons are emitted and photocurrent starts flowing. But the number of photoelectrons ejected depends on the intensity of incident light, higher the intensity, higher will be the electrons ejected resulting in increase in photoelectric current.

Green light ejects electron from a certain photo-sensitive surface, but yellow light does not. What will happen in case of red and violet light?

The frequency of the red light photon is less than the threshold frequency of the photo metal, whereas the frequency of violet light photon is greater than the threshold frequency of the photo metal. Thus, in case of red light photons no electrons will be ejected whereas in the case of violet light photons the electrons will be ejected out of the photo metal.

Why does photoelectric emission does not take place if the frequency of incident radiation is less than threshold frequency?

According to Einstein's photoelectric equation : Maximum kinetic energy K m a x = h ν - h ν 0 Since K m a x is positive. So, the photoelectric emission occurs only if h ν > h ν 0 ⇒ ν > ν 0 Therefore, the photoelectric emission occurs only when the incident light has more than a certain minimum frequency. This minimum frequency is called threshold frequency.

Derive de Broglie wave equation for material particles.

According to Plank's theory,the energy associated with a photon is given by E = hν where ν = frequency and h = Plank's constant According to the Einstein theory of relativity, the same quantity is related to its mass and the velocity by the relation E = mc 2 Thus, energy can be given in terms of frequency (a wave characteristic) as well in terms of mass (a particle property) as per the principle of dual nature of matter, the two forms can be unified, hence de-Broglie equated both these equations for the particle moving with velocity v as m v 2 = hν = hc λ where λ is the wavelength of the particle From this h λ = mv ⇒ λ = h mv = h P Where P is the momentum of particle and h is the Plank's constant. This equation relating the momentum of a particle with its wavelength is the de Broglie equation.

How does the maximum kinetic energy of electron vary with work function of metal?

The maximum kinetic energy of the electron E K = h ν - W where W is the work function. From the equation of work function, the smaller the work function W , the greater the kinetic energy. This means that when the work function of metal increases, the maximum kinetic energy of photoelectrons decreases.