Wavelength And Frequency

We are given that two materials, A and B, have work functions of 3 eV and 4 eV, respectively. Both emit photoelectrons with the same maximum kinetic energy of 1 eV. We are also told that the wavelength of incident light on material A is 500 nm. Our task is to determine the wavelength of the incident light on material B.

The energy of a photon is related to its wavelength by the equation:

Where: - is Planck's constant ( ), - is the speed of light ( ), - is the wavelength of the light.

The photoelectric equation is:

Where: - is the work function of the material, - is the maximum kinetic energy of the emitted photoelectrons. For material A, the work function is 3 eV, and the maximum kinetic energy of the photoelectrons is 1 eV. Using the given wavelength , we calculate the energy of the photon incident on A: Convert the wavelength to meters: . Using and , we can calculate the photon energy in joules and then convert it to electron volts (1 eV = J). For material B, we use the same photoelectric equation with the work function and the same maximum kinetic energy . Using the photoelectric equation for B: Now, using the energy equation for material B: Solving for : Substituting the values:

Correct Answer:

Correct Answer: (C) 350 nm

The de-Broglie wavelength ( ) of a particle is given by the equation:

Where: - is Planck's constant, - is the momentum of the particle.

We are given that the momentum of the α-particle is half that of the proton. Let's denote the momentum of the proton as and the momentum of the α-particle as .

The de-Broglie wavelength for the proton is:

The de-Broglie wavelength for the α-particle is:

Thus, the ratio of the wavelengths of the de-Broglie waves of the α-particle and the proton is:

Correct Answer:

Correct Answer: (D) 2:1

We need to pick out the incorrect statement from the following options:

• Speech signal requires a bandwidth of 2800 Hz: This is correct. Speech signals typically require a bandwidth of around 2800 Hz.
• The approximate bandwidth to transmit music is 20 kHz: This is correct. Music signals generally require a bandwidth of about 20 kHz for good quality transmission.
• The bandwidth of video signals required to transmit pictures is 4.2 MHz: This is correct. A typical bandwidth required for transmitting video signals, especially for television broadcasts, is around 4.2 MHz.
• The bandwidth usually allocated to transmit TV signals is 6 MHz: This statement is incorrect. In most countries, the bandwidth allocated for TV signals (analog) is 6 MHz for each channel in the U.S. and 8 MHz in some other countries like Japan. However, this is not universally fixed at 6 MHz for all TV signals worldwide, especially considering digital TV broadcasting.
• Digital signals are usually in the form of sine waves: This is incorrect. Digital signals are typically represented by square waves, not sine waves.

Correct Answer: (D) The bandwidth usually allocated to transmit TV signals is 6 MHz

In a transmitter, the audio signal of frequency is modulated by the carrier signal of frequency . The modulation process generates a signal that contains three components:

• The carrier frequency ,
• The upper sideband ,
• The lower sideband .

The band-pass filter in the transmitter is used to select only the frequencies around the carrier frequency and reject the unwanted frequencies. This means that the filter will reject:

• The carrier frequency ,
• The audio signal frequency ,
• The frequencies and , which correspond to the lower and upper sidebands of the modulated signal.

Thus, the band-pass filter will reject the frequencies and , as those are not part of the intended modulated signal.

Correct Answer:

Correct Answer: (E) and

The time period is the time taken for one complete cycle. Given that the musician hits the drum 90 times in a minute (60 seconds), we can find the time period by: Since there are 60 seconds in a minute:
Thus, the time period is .